// Numbas version: exam_results_page_options {"name": "Maths Support: Maxima and minima for differentiable functions on intervals", "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "duration": 0.0, "metadata": {"notes": "\n \t\t

Finding global maxima and global minima over open and closed intervals

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\n \t\t", "description": "

5 questions on finding local and global maxima and minima on compact intervals and on the real line for differentiable functions.

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Differentiating, we have:

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\\[g'(x)=\\simplify{{c}*x^2+{-c*(a+b)}*x+{c*a*b}={c}*(x+{-a})(x-{b})}\\]

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Note that we have already factorised the derivative.

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Stationary points are given by solving $g'(x)=0 \\Rightarrow x=\\var{a},\\;\\;\\mbox{or }x=\\var{b}$

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So the least stationary point is $x=\\var{a}$ and the greatest is $x=\\var{b}$.

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Since $\\var{a} > \\var{l}$ and $\\var{b} \\lt \\var{m}$ we have that both stationary points are in $I$.

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The second derivative is given by \\[g''(x)=\\simplify{{2*c}*x-{c*(a+b)}}\\]

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Local Maximum

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At the stationary point $x=\\var{a}$ we have $g''(\\var{a})=\\var{c*a-c*b} \\lt 0$.

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Hence at this value of $x$ we have a local maximum.

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The value of the function $g$ at this local maximum is $g(\\var{a})= \\var{valmax}$.

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Local Minimum

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At the stationary point $x=\\var{b}$ we have $g''(\\var{b})=\\var{c*b-c*a} \\gt 0$.

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Hence this point is a local minimum.

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The value of the function $g$ at this local minimum is $g(\\var{b})= \\var{valmin}$.

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Global Maximum

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First we find the values at the endpoints of the interval $I=[\\var{l},\\var{m}]$ are:

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$g(\\var{l})=\\var{valbegin}$ to 3 decimal places.

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$g(\\var{m})=\\var{valend}$ to 3 decimal places.

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To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$.

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So we proceed by comparing the values of the function at the endpoints with the local maximum.

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a) If the value at the local maximum is greater than either of the values at the endpoints then this is the global maximum on the interval.

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b) Otherwise if the greatest value of the function at the endpoints is greater than the local maximum then this is the global maximum.

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\\[\\begin{array}{c|c|c|c} x & \\mbox{Local Maximum}=\\var{a} & \\var{l} \\in I & \\var{m} \\in I \\\\ \\hline\\\\ g(x)& \\var{valmax} & \\var{valbegin} & \\var{valend} \\\\ \\end{array} \\]

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So for our example we see that the global maximum occurs at $x=\\var{gma}$ and $g(\\var{gma})=\\var{valgmax}$.

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Global Minimum

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We proceed as for the global maximum by comparing the values of the function at the endpoints with the local minimum.

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a) If the value at the local minimum is less than either of the values at the endpoints then this is the global minimum on the interval.

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b) Otherwise if the least value of the function at the endpoints is less than the local minimum then this is the global minimum.

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\\[\\begin{array}{c|c|c|c} x & \\mbox{Local Minimum}=\\var{b} & \\var{l} \\in I & \\var{m} \\in I \\\\ \\hline\\\\ g(x)& \\var{valmin} & \\var{valbegin} & \\var{valend} \\\\ \\end{array} \\]

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In our example we see that the global minimum occurs at $x=\\var{gmi}$ and $g(\\var{gmi})=\\var{valgmin}$.

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Input the first derivative of $g$ here, factorised into a product of two linear factors in the form $g'(x)=c(x-a)(x-b)$for suitable integers $a$, $b$ and $c$:

\n \n

$g'(x)=\\;\\;$[[0]]

\n \n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Factorise the expression

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Factorise the expression

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

Find the stationary points of $g$.

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Least stationary point: [[0]]    Greatest stationary point: [[1]]

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Do both these stationary points lie in the interval $I$ ? [[2]]

\n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"displayColumns": 0, "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

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No

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Input the second derivative of $g$:

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$g''(x)=\\;\\;$ [[0]]

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Hence find all local maxima and minima given by the stationary points

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Local maximum is at $x=\\;\\;$ [[1]] and the value of the function at the local maximum = [[2]]

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Local minimum is at $x=\\;\\;$ [[3]] and the value of the function at the local minimum = [[4]]

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What are the following values at the end points of the interval $I$ ?

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$g(\\var{l})=\\;\\;$ [[0]]      $g(\\var{m})=\\;\\;$ [[1]]

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Input both to 3 decimal places.

\n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "valbegin", "minValue": "valbegin", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "valend", "minValue": "valend", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

Global Maximum

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At what value of $x \\in I$ does $g$ have a global maximum ?

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$x=\\;\\;$ [[0]]

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Value of $g$ at this global maximum = [[1]] (input to 3 decimal places).

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Global Minimum

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At what value of $x \\in I$ does $g$ have a global minimum ?

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$x=\\;\\;$ [[2]]

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Value of $g$ at this global minimum = [[3]] (input to 3 decimal places).

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Let $I=[\\var{l},\\var{m}]$ be an interval and let $g: I \\rightarrow I$ be a function defined on this interval
given by :\\[g(x) = \\simplify{{c}/3*x^3+ {-c*(a+b)}/2*x^2+{c*a*b}*x+{d}}\\]

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9/07/2102:

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Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t

Changed grammar in the Advice section.

\n \t\t", "description": "

$I$ compact interval, $g:I\\rightarrow I,\\;g(x)=ax^3+bx^2+cx+d$. Find stationary points, local and global maxima and minima of $g$ on $I$

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First derivative

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Differentiating we have:

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\\[\\begin{eqnarray*} g'(x)&=&\\simplify{(x-{b})^3+3*(x-{a})*(x-{b})^2}\\\\ &=&\\simplify{(x-{b})^2(3*(x-{a})+x-{b})}\\\\ &=&\\simplify{4*(x-{k})*(x-{b})^2} \\end{eqnarray*} \\] and we have factorised the expression.

\n

Stationary points

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These are given by solving $g'(x)=0 \\Rightarrow x=\\var{k},\\;\\;\\mbox{or }x=\\var{b}$

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Therefore the least stationary point is $x=\\var{k}$ and the greatest is $x=\\var{b}$ and we see that both stationary points are in $I$.

\n

Second derivative.

\n

The second derivative is given by:
\\[\\begin{eqnarray*} g''(x)&=&\\simplify{4*(x-{b})^2+8*(x-{k})(x-{b})}\\\\ &=&\\simplify{4*(x-{b})(3*x-{2*k+b})} \\end{eqnarray*} \\]

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Local Minimum

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At the stationary point $x=\\var{k}$ we have $g''(\\var{k})=\\var{4*(k-b)^2} \\gt 0$.

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Hence $x=\\var{k}$ is a local minimum.

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The other stationary point

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The value at $x=\\var{b}$ is $g(\\var{b})= 0$.

\n

Hence this test fails at this point and we proceed to use the third derivative to see in more information can be gained.

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Third derivative

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We see that $g'''(x)=\\simplify{8*(3*x-{k+2*b})}$.

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Testing the stationary point using the third derivative gives:

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$g'''(\\var{b})=\\var{8*(b-k)} \\neq 0$.

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Therefore there cannot be an extremum point at $x=\\var{b}$.

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Finding the global maximum and minimum on $I$

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First we find the values at the endpoints of the interval $I=[\\var{l},\\var{m}]$ are:

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$g(\\var{l})=\\var{valbegin}$.

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$g(\\var{m})=\\var{valend}$.

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Global Maximum

\n

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$ and since $g$ does not have a local maximum on $I$ it must take its maximum value at one of the end points of $I$.

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We see from the values at the end points obtained above that the global maximum value on $I$ is at $x=\\var{xma}$.

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We have $g(\\var{xma})=\\var{gma}$.

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Global Minimum.

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$g$ has only one local minimum on $I$ at $x=\\var{k}$ and so this must be the global minimum on $I$.

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We have $g(\\var{k})=\\var{(k-a)*(k-b)^3}$.

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n

Input the first derivative of $g$ here, factorised into a product of two factors in the form $g'(x)=c(x-a)(x-b)^2$for suitable integers $a$, $b$ and $c$:

\n \n

$g'(x)=\\;\\;$[[0]]

\n \n ", "gaps": [{"notallowed": {"message": "

Factorise the expression

", "showstrings": false, "strings": ["x^2", "x^3"], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "4*(x + {-k}) * (x + {-b})^2", "type": "jme", "musthave": {"message": "

Factorise the expression

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Find the stationary points of $g$, here thought of as $g:\\mathbb{R} \\rightarrow \\mathbb{R}$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]]

\n \n

Do both these stationary points lie in the interval $I$ ? [[2]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{k}", "type": "jme"}, {"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{b}", "type": "jme"}, {"maxanswers": 0.0, "matrix": [1.0, 0.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

Yes

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No

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Input the second derivative of $g$:

\n \n

$g''(x)=\\;\\;$ [[0]]

\n \n

Using $g''(x)$, determine more information about the stationary points:

\n \n

Least stationary point is: (Choose one of the following)
[[1]]

\n \n

Greatest stationary point is: (Choose one of the following)
[[2]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "4*(x-{b})*(3*x-{2*k+b})", "type": "jme"}, {"maxanswers": 0.0, "matrix": [1.0, 0.0, 0.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

A local minimum.

", "

A local maximum.

", "

Uncertain as the second derivative test fails.

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A local minimum.

", "

A local maximum.

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Uncertain as the second derivative test fails.

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Using the third derivative answer the following questions:
$g'''(x) = \\;\\;$[[0]]

\n \n

If $a$ is the least stationary point then $g'''(a) =\\;\\;$[[1]]

\n \n

If $b$ is the other stationary point then $g'''(b) =\\;\\;$[[2]]

\n \n

This information tells us that: (Choose one of the following).
[[3]]

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{k} is not a local minimum or a local maximum.

", "

{b} is not a local minimum or a local maximum.

", "

{b} is not a local minimum or a local maximum and neither is {k}.

", "

{k} is a local maximum and {b} is a local minimum.

", "

{k} is a local minimum and {b} is a local maximum.

"], "marks": 0.0, "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", "", "", ""], "displaycolumns": 0.0, "type": "1_n_2", "minmarks": 0.0}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

What are the following values at the end points of the interval $I$ ?

\n \n

$g(\\var{l})=\\;\\;$ [[0]]

\n \n

$g(\\var{m})=\\;\\;$ [[1]]

\n \n

Input both to 2 decimal places.

\n \n ", "gaps": [{"minvalue": "{valbegin}", "type": "numberentry", "maxvalue": "{valbegin}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{valend}", "type": "numberentry", "maxvalue": "{valend}", "marks": 0.5, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

Global Maximum

\n \n

At what value of $x \\in I$ does $g$ have a global maximum ?

\n \n

$x=\\;\\;$ [[0]]

\n \n

Value of $g$ at this global maximum = [[1]].

\n \n

Global Minimum

\n \n

At what value of $x \\in I$ does $g$ have a global minimum ?

\n \n

$x=\\;\\;$ [[2]]

\n \n

Value of $g$ at this global minimum = [[3]].

\n \n ", "gaps": [{"minvalue": "{xma}", "type": "numberentry", "maxvalue": "{xma}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{gma}", "type": "numberentry", "maxvalue": "{gma}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{k}", "type": "numberentry", "maxvalue": "{k}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{gmi}", "type": "numberentry", "maxvalue": "{gmi}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

Let $I=[\\var{l},\\var{m}]$ be an interval and let $g: I \\rightarrow \\mathbb{R}$ be a function defined on this interval
given by :\\[g(x) = \\simplify{(x-{a})*(x-{b})^3}\\]

\n \n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "s1*random(1..9)", "name": "a"}, "c": {"definition": "random(3,6)", "name": "c"}, "b": {"definition": "4*k-3*a", "name": "b"}, "d": {"definition": "if(c=3,random(-9..9),random(-9..9#2))", "name": "d"}, "gma": {"definition": "(xma-a)*(xma-b)^3", "name": "gma"}, "k": {"definition": "a+random(1..6)", "name": "k"}, "m": {"definition": "b+random(1..3)", "name": "m"}, "xma": {"definition": "if(valend > valbegin,m,l)", "name": "xma"}, "l": {"definition": "k-random(1..3)", "name": "l"}, "valend": {"definition": "(m-a)*(m-b)^3", "name": "valend"}, "gmi": {"definition": "(k-a)*(k-b)^3", "name": "gmi"}, "valbegin": {"definition": "(l-a)*(l-b)^3", "name": "valbegin"}, "third2": {"definition": "8*(b-k)", "name": "third2"}, "third1": {"definition": "16*(k-b)", "name": "third1"}, "s1": {"definition": "random(1,-1)", "name": "s1"}}, "metadata": {"notes": "\n \t\t

9/07/2012:

\n \t\t

Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t", "description": "

$I$ compact interval, $g:I\\rightarrow I$, $g(x)=(x-a)(x-b)^2$. Stationary points in interval. Find local and global maxima and minima of $g$ on $I$. 

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Max and Min 4", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "b", "valmin", "valmax", "lmi", "s", "tol", "lma"], "tags": ["Calculus", "calculus", "classifying stationary points", "functions", "global maxima and minima", "limit", "limits", "local maxima and minima", "maxima and minima", "optimisation", "optimising", "optimising functions", "rational polynomials", "stationary points", "taking limits"], "preamble": {"css": "", "js": ""}, "advice": "\n

The function $g(x)$ is continuous and differentiable at all points in $\\mathbb{R}$.

\n

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g'(x)&=&\\simplify{({a}*(x^2+{b^2})-{2*a}*x^2)/(x^2+{b^2})^2}\\\\ &=&\\simplify{({-a}*(x-{b})(x+{b}))/(x^2+{b^2})^2} \\end{eqnarray*} \\]

\n

Stationary Points.

\n

The stationary points are given by solving $g'(x)=0$.

\n

$g'(x)=0 \\Rightarrow \\simplify{{-a}*(x-{b})(x+{b})=0} \\Rightarrow x=\\var{b} \\mbox{ or } x=\\var{-b}$

\n

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

\n

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g''(x)&=&\\simplify[std]{({-2*a}*x*(x^2+{b^2})^2+{4*a}*x*(x^2-{b^2})(x^2+{b^2}))/(x^2+{b^2})^4}\\\\ &=&\\simplify[std]{({2*a}*x*(x^2-{3*b^2}))/(x^2+{b^2})^3} \\end{eqnarray*} \\]

\n

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

\n

For $x= \\var{lma}$ we have: \\[g''(\\var{lma})= \\simplify[std]{-{abs(a)}/{2*b^3}} \\lt 0\\]

\n

Hence is a local maximum.

\n

Evaluating the function at $x=\\var{lma}$ gives $g(\\var{lma})=\\var{valmax}$ to 3 decimal places.

\n

For $x= \\var{lmi}$ we have: \\[g''(\\var{lmi})= \\simplify[std]{{abs(a)}/{2*b^3}} \\gt 0\\]

\n

Hence is a local minimum.

\n

Evaluating the function at $x=\\var{lmi}$ gives $g(\\var{lmi})=\\var{valmin}$ to 3 decimal places.

\n

The Limits.

\n

If we divide $g(x)$ top and bottom by $x^2$ (OK as $x \\neq 0$ at any time) we obtain: \\[g(x)=\\simplify[std]{({a}/x)/(1+{b^2}/x^2)}\\]

\n

Then using the fact that $\\displaystyle \\frac{1}{x}$ and $\\displaystyle \\frac{1}{x^2}$ both tend to $0$ as $ x \\rightarrow \\pm\\infty$ we see that

\n

$\\displaystyle \\lim_{x \\to \\infty}g(x)=\\frac{0}{1}=0$ and similarly

\n

$\\lim_{x \\to -\\infty}g(x)=0$

\n

Global Maximum and Minimum

\n

Since $g$ has a finite limit of $0$ as $x \\rightarrow \\pm\\infty$ and we have that $0$ lies between the local minimum $\\var{valmin}$ and the local maximum $\\var{valmax}$

\n

Then:

\n

Global Maximum: The local maximum of $g$ we have found at $x=\\var{lma}$ must be a global maximum and similarly,

\n

Global Minimum: The local minimum of $g$ we have found at $x=\\var{lmi}$ must be a global minimum.

\n

So we have shown \\[\\forall x \\in \\mathbb{R},\\;\\;\\var{valmin} \\le g(x) \\le \\var{valmax}\\]

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"displayColumns": 0, "prompt": "\n

Is $g(x)$ continuous at all points of $\\mathbb{R}$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": false, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

The first derivative of $g$ can be written in the form $\\displaystyle \\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)=(x^2+\\var{b^2})^2$ are polynomials.

\n

Input the numerator $p(x)$ of the first derivative of $g$ here, factorised into a product of two linear factors in the form
\\[p(x)=c(x-a)(x-b)\\]for suitable integers $a$, $b$ and $c$:

\n

$p(x)\\;=\\;$[[0]]

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["^", "x*x", "xx", "x x"], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "({( - a)} * (x + ( - {b})) * (x + {b}))", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme", "musthave": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"displayColumns": 0, "prompt": "\n

Is $g(x)$ differentiable at all points of $\\mathbb{R}$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

Find the stationary points of $g$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]]

\n \n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{-b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

The second derivative of $g$ can be written in the form $\\displaystyle \\frac{r(x)}{s(x)}$ where $r(x)$ and $s(x)=(x^2+\\var{b^2})^3$ are polynomials.

\n

Input the numerator $r(x)$ of the second derivative of $g$ here, factorised into a product of a linear factor and a quadratic factor in the form
\\[r(x)=a_1x(x^2-a_2)\\] for suitable integers $a_1$, $a_2$

\n

$r(x)=\\;\\;$ [[0]]

\n

Hence find all local maxima and minima given by the stationary points

\n

Local maximum is at $x=\\;\\;$ [[1]] and the value of the function at the local maximum (to 3 decimal places)= [[2]]

\n

Local minimum is at $x=\\;\\;$ [[3]] and the value of the function at the local minimum (to 3 decimal places) = [[4]]

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Factorise the expression as asked in the question.

", "showStrings": false, "strings": ["x^3"], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{2*a}*x*(x^2-{3*b^2})", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme", "musthave": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{lma}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"allowFractions": false, "marks": 1, "maxValue": "{valmax+tol}", "minValue": "{valmax-tol}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{lmi}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"allowFractions": false, "marks": 1, "maxValue": "{valmin+tol}", "minValue": "{valmin-tol}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

What are the following limits?

\n

1) $\\lim_{x \\to \\infty}g(x)\\;\\;$

\n

Choose one of the following [[0]]

\n

2) $\\lim_{x \\to -\\infty}g(x)$

\n

Choose one of the following [[1]]

\n

Does $g$ have a finite global maximum? Click on Yes or No
[[2]]

\n

Does $g$ have a finite global maximum? Click on Yes or No
[[2]]

\n ", "marks": 0, "gaps": [{"displayColumns": 0, "matrix": [0, 0, 0, 0, 1], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", "", "", "", ""], "choices": ["

$-\\infty$

", "

$\\infty$

", "

$\\var{b}$

", "

$\\var{valmax}$

", "

$0$

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"displayColumns": 0, "matrix": [0, 0, 0, 0, 1], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", "", "", "", ""], "choices": ["

$-\\infty$

", "

$\\infty$

", "

$\\var{a}$

", "

$\\var{valmin}$

", "

$0$

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"displayColumns": 0, "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"displayColumns": 0, "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "

Let $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ be the function given by:
\\[g(x)=\\simplify{{a}*x/(x^2+{b}^2)}\\]

", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "s*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "b": {"definition": "random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "valmin": {"definition": "precround(-abs(a)*b/(2*b^2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valmin", "description": ""}, "valmax": {"definition": "-valmin", "templateType": "anything", "group": "Ungrouped variables", "name": "valmax", "description": ""}, "lmi": {"definition": "if(a<0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lmi", "description": ""}, "s": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s", "description": ""}, "tol": {"definition": "0.001", "templateType": "anything", "group": "Ungrouped variables", "name": "tol", "description": ""}, "lma": {"definition": "if(a>0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lma", "description": ""}}, "metadata": {"notes": "\n \t\t

10/07/2012:

\n \t\t

Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t

9/07/2012:

\n \t\t

Added tags.

\n \t\t

Corrected mistake in Advice ($x$ instead of $x^2$).

\n \t\t

Tolerance variable set to tol=0.001 for a numeric entry.

\n \t\t", "description": "

$g: \\mathbb{R} \\rightarrow \\mathbb{R}, g(x)=\\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min? 

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Max and Min 5", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "valmax", "s", "lmi", "valend", "tol", "valbegin", "lma", "valmin"], "tags": ["Calculus", "calculus", "classifying stationary points", "finding local maxima and minima", "finding the stationary points", "limit", "limits", "maxima and minima", "maximum", "min", "minimum", "optimisation", "optimising functions on an interval", "stationary points", "taking limits"], "preamble": {"css": "", "js": ""}, "advice": "

The function $g(x)$ is continuous and differentiable at all points in $\\mathbb{R}$.

\n

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g'(x)&=&\\simplify{({a}*(x^2+{b^2})-{2*a}*x^2)/(x^2+{b^2})^2}\\\\ &=&\\simplify{({-a}*(x-{b})(x+{b}))/(x^2+{b^2})^2} \\end{eqnarray*} \\]

\n

Stationary Points.

\n

The stationary points are given by solving $g'(x)=0$.

\n

$g'(x)=0 \\Rightarrow \\simplify{{-a}*(x-{b})(x+{b})=0} \\Rightarrow x=\\var{b} \\mbox{ or } x=\\var{-b}$

\n

We see that both stationary points are in the inerval $I$.

\n

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

\n

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g''(x)&=&\\simplify[std]{({-2*a}*x*(x^2+{b^2})^2+{4*a}*x*(x^2-{b^2})(x^2+{b^2}))/(x^2+{b^2})^4}\\\\ &=&\\simplify[std]{({2*a}*x*(x^2-{3*b^2}))/(x^2+{b^2})^3} \\end{eqnarray*} \\]

\n

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

\n

For $x= \\var{lma}$ we have: \\[g''(\\var{lma})= \\simplify[std]{-{abs(a)}/{2*b^3}} \\lt 0\\]

\n

Hence is a local maximum.

\n

Evaluating the function at $x=\\var{lma}$ gives $g(\\var{lma})=\\var{valmax}$ to 3 decimal places.

\n

For $x= \\var{lmi}$ we have: \\[g''(\\var{lmi})= \\simplify[std]{{abs(a)}/{2*b^3}} \\gt 0\\]

\n

Hence is a local minimum.

\n

Evaluating the function at $x=\\var{lmi}$ gives $g(\\var{lmi})=\\var{valmin}$ to 3 decimal places.

\n

The values of $g$ at the endpoints are:

\n

$g(\\var{c})=\\var{valbegin}$ and $g(\\var{d})=\\var{valend}$ to 3 decimal places.

\n

Global Maximum and Minimum

\n

Since $g$ has a finite limit of $0$ as $x \\rightarrow \\pm\\infty$ and we have that $0$ lies between the local minimum value $\\var{valmin}$ and the local maximum value $\\var{valmax}$ (and these occur at values in $I$).

\n

then:

\n

Global Maximum: The local maximum of $g$ we have found at $x=\\var{lma} \\in I$ must be a global maximum and similarly,

\n

Global Minimum: The local minimum of $g$ we have found at $x=\\var{lmi} \\in I$ must be a global minimum.

\n

So we have shown \\[\\forall x \\in \\mathbb{R},\\;\\;\\var{valmin} \\le g(x) \\le \\var{valmax}\\]

\n

(all to 3 decimal places).

", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"displayColumns": 0, "prompt": "\n

Is $g(x)$ continuous at all points of $I$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": false, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

The first derivative of $g$ can be written in the form $\\displaystyle \\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)=(x^2+\\var{b^2})^2$ are polynomials.

\n

Input the numerator $p(x)$ of the first derivative of $g$ here, factorised into a product of two linear factors in the form
\\[p(x)=c(x-a)(x-b)\\]for suitable integers $a$, $b$ and $c$:

\n

$p(x)=\\;\\;$[[0]]

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["^", "x*x", "xx", "x x"], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "({( - a)} * (x + ( - {b})) * (x + {b}))", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme", "musthave": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"displayColumns": 0, "prompt": "\n

Is $g(x)$ differentiable at all points of $I$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

Assume now that $g$ is a function $g:\\mathbb{R} \\rightarrow \\mathbb{R}$.

\n \n

Find the stationary points of $g$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]]

\n \n

Are both stationary points in the interval $I$? Choose Yes or No.
[[2]]

\n \n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{-b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"displayColumns": 0, "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

The second derivative of $g$ can be written in the form $\\displaystyle \\frac{r(x)}{s(x)}$ where $r(x)$ and $s(x)=(x^2+\\var{b^2})^3$ are polynomials.

\n

Input the numerator $r(x)$ of the second derivative of $g$ here, factorised into a product of a linear factor and a quadratic factor in the form
\\[r(x)=a_1x(x^2-a_2)\\] for suitable integers $a_1$, $a_2$

\n

$r(x)=\\;\\;$ [[0]]

\n

Hence find all local maxima and minima given by the stationary points

\n

Local maximum is at $x=\\;\\;$ [[1]] and the value of the function at the local maximum (to 3 decimal places)= [[2]]

\n

Local minimum is at $x=\\;\\;$ [[3]] and the value of the function at the local minimum (to 3 decimal places) = [[4]]

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["x^2", "x x", "xx"], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{2*a}*x*(x^2-{3*b^2})", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme", "musthave": {"message": "

Factorise the expression

", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{lma}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"allowFractions": false, "marks": 1, "maxValue": "{valmax+tol}", "minValue": "{valmax-tol}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{lmi}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"allowFractions": false, "marks": 1, "maxValue": "{valmin+tol}", "minValue": "{valmin-tol}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

What are the following values at the end points of the interval $I$ ?

\n \n

$g(\\var{c})=\\;\\;$ [[0]]

\n \n

$g(\\var{d})=\\;\\;$ [[1]]

\n \n

Input both to 3 decimal places.

\n \n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "precround(valbegin+0.001,3)", "minValue": "precround(valbegin-0.001,3)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "precround(valend+0.001,3)", "minValue": "precround(valend-0.001,3)", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

Assume now that $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ and you are given that:

\n \n

$\\lim_{x \\to \\infty}g(x)=0$ and $\\lim_{x \\to -\\infty}g(x)=0$

\n \n

Global Maximum

\n \n

At what value of $x \\in I$ does $g$ have a global maximum ?

\n \n

$x=\\;\\;$ [[0]]

\n \n

Global Minimum

\n \n

At what value of $x \\in I$ does $g$ have a global minimum ?

\n \n

$x=\\;\\;$ [[1]]

\n \n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "{lma}", "minValue": "{lma}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "{lmi}", "minValue": "{lmi}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "\n

Let $I=[\\var{c},\\var{d}]$ be an interval and let $g: I \\rightarrow I$ be the function given by:
\\[g(x)=\\simplify{{a}*x/(x^2+{b}^2)}\\]

\n \n

Answer the following questions. There are seven parts and you may need to scroll down to complete all parts.

\n \n ", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "s*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "-b-random(1..3)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "b+random(1..3)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "valmax": {"definition": "-valmin", "templateType": "anything", "group": "Ungrouped variables", "name": "valmax", "description": ""}, "valend": {"definition": "precround(a*d/(d^2+b^2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valend", "description": ""}, "lmi": {"definition": "if(a<0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lmi", "description": ""}, "s": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s", "description": ""}, "tol": {"definition": "0.001", "templateType": "anything", "group": "Ungrouped variables", "name": "tol", "description": ""}, "valbegin": {"definition": "precround(a*c/(c^2+b^2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valbegin", "description": ""}, "lma": {"definition": "if(a>0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lma", "description": ""}, "valmin": {"definition": "precround(-abs(a)*b/(2*b^2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valmin", "description": ""}}, "metadata": {"notes": "\n \t\t

9/07/2012:

\n \t\t

Added tags.

\n \t\t

Corrected mistake in Advice ($x$ instead of $x^2$).

\n \t\t

Tolerance variable set to tol=0.001 for a numeric entry.

\n \t\t

 

\n \t\t

10/07/2012:

\n \t\t

Added tags.

\n \t\t

Edited grammar in the Advice section.

\n \t\t

Question appears to be working correctly.

\n \t\t", "description": "

$I$ compact interval. $\\displaystyle g: I \\rightarrow I, g(x)=\\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min? 

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Max and Min 6", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "valmin", "m", "l", "xma", "lmi", "s", "u", "valend", "tol", "l1", "valbegin", "lma", "statpoint", "xmi", "valsd"], "tags": ["Calculus", "Differentiation", "calculus", "classifying stationary points", "continuous functions", "derivative", "derivatives", "differentiable functions", "differentiate", "differentiation", "first derivative", "functions", "global maximum", "global minimum", "local maximum", "local minimum", "max and min", "maxima", "maximum", "maximum and minimum", "maximum and minimum of a function on an interval", "minima", "minimum", "optimising a function on an interval", "quadratic", "quadratics", "quotient rule", "second derivative", "solution of a quadratic", "stationary points"], "preamble": {"css": "", "js": ""}, "advice": "\n

The function $g(x)$ is continuous and differentiable at all points in $\\mathbb{R}$ as $ \\var{c} \\notin I$.

\n

On differentiating we see that
\\[\\begin{eqnarray*}g'(x)&=&\\simplify{2*x/(x-{c})^({a}/{b}) - {a}*x^2/({b}(x-{c})^({a+b}/{b}))}\\\\ &=&\\simplify{({2*b}x*(x-{c})-{a}*x^2)/({b}(x-{c})^({a+b}/{b}))}\\\\ &=&\\simplify{(x*({2*b-a}x-{2*b*c}))/({b}(x-{c})^({a+b}/{b}))} \\end{eqnarray*} \\]

\n

Stationary Points.

\n

The stationary points are given by solving $g'(x)=0$.

\n

$\\displaystyle g'(x)=0 \\Rightarrow \\simplify{x*({2*b-a}x-{2*b*c})=0} \\Rightarrow x=0 \\mbox{ or } x=\\simplify[std]{{2*b*c}/{2*b-a}}$

\n

We see that $\\displaystyle x=\\simplify[std]{{2*b*c}/{2*b-a}}$ is the only stationary point in $I$.

\n

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

\n

Using the quotient rule for differentiation we see that
\\[g''(x)=\\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})/({b^2}(x-{c})^({a+2*b}/{b}))}\\]

\n

Hence \\[r(x)=\\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})}\\]

\n

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

\n

There is only one stationary point $\\displaystyle  x=\\simplify[std]{{2*b*c}/{2*b-a}}$ in $I$ and at that point we have:
\\[g''\\left(\\simplify[std]{{2*b*c}/{2*b-a}}\\right)=\\var{valsd} \\gt 0\\]

\n

Hence this point is a local minimum.

\n

Evaluating at end points of the interval.

\n

The values of $g$ at the endpoints are:

\n

$g(\\var{l})=\\var{valbegin}$ and $g(\\var{m})=\\var{valend}$, both to 3 decimal places.

\n

Global Maximum and Minimum

\n

Global Maximum: Since $g$ does not have a local maximum in the interval $I$, it must take a global maximum value at one of the end points of the interval.

\n

From the values found for $g(\\var{l})$ and $g(\\var{m})$ found above, we see that $x=\\var{xma}$ is the global maximum for $g$ in the interval $I$.

\n

Global Minimum: The local minimum of $g$ given by $ \\displaystyle x=\\simplify[std]{{2*b*c}/{2*b-a}} \\in I$ is the only local minimum and must be a global minimum in $I$.

\n

Note that the global minimum value for $g$ on $I$ is:

\n

\\[g\\left(\\simplify[std]{{2*b*c}/{2*b-a}}\\right)=\\var{valmin}\\]

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"displayColumns": 0, "prompt": "\n

Is $g(x)$ continuous at all points of $I$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": false, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

The first derivative of $g$ can be written in the form $\\displaystyle \\frac{p(x)}{q(x)}$ where $p(x)$ is a polynomial of degree $2$ and $q(x)=\\simplify{{b}*(x-{c})^({a+b}/{b})}$.

\n

Input the numerator $p(x)$ of the first derivative of $g$ here:

\n

$p(x)=\\;\\;$[[0]]

\n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "x*({2*b-a}*x-{2*b*c})", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"displayColumns": 0, "prompt": "\n

Is $g(x)$ differentiable at all interior points of $I$?

\n \n

Choose Yes or No.

\n \n ", "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Yes

", "

No

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}, {"prompt": "\n

Assume now that $g$ is a function $g:\\mathbb{R} \\backslash \\{\\var{c}\\} \\rightarrow \\mathbb{R}$.

\n \n

Find the stationary points of $g$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]] (Input as a fraction or an integer and not as a decimal)

\n \n

Which stationary point is in the interval $I$? Choose one of the following:
[[2]]

\n \n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{0}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"notallowed": {"message": "

Input as a fraction or an integer and not as a decimal

", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{2*b*c}/{2*b-a}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"displayColumns": 0, "matrix": [1, 0], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

The greatest

", "

The least

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

The second derivative of $g$ can be written in the form $\\displaystyle \\frac{r(x)}{s(x)}$ where $r(x)$ is a quadratic polynomial and $s(x)=\\simplify{{b^2}(x-{c})^({a+2*b}/{b})}$.

\n

Input the numerator $r(x)$ of the second derivative of $g$ here:

\n

$r(x)=\\;\\;$ [[0]]

\n

Hence determine the type of the stationary point which lies in $I$. Choose one of the following:
[[1]]

\n ", "marks": 0, "gaps": [{"expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"displayColumns": 0, "matrix": [0, 1], "shuffleChoices": true, "maxMarks": 0, "distractors": ["", ""], "choices": ["

Local maximum

", "

Local minimum

"], "displayType": "radiogroup", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "1_n_2", "minMarks": 0}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

What are the following values at the end points of the interval $I$ ?

\n \n

$g(\\var{l})=\\;\\;$ [[0]]

\n \n

$g(\\var{m})=\\;\\;$ [[1]]

\n \n

Input both to 3 decimal places.

\n \n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "valbegin+tol", "minValue": "valbegin-tol", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "valend+tol", "minValue": "valend-tol", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

Global Maximum

\n \n

At what value of $x \\in I$ does $g$ have a global maximum in $I$?

\n \n

$x=\\;\\;$ [[0]]

\n \n

Global Minimum

\n \n

At what value of $x \\in I$ does $g$ have a global minimum in $I$ ?

\n \n

$x=\\;\\;$ [[1]] (Input as a fraction or an integer and not as a decimal)

\n \n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "{xma}", "minValue": "{xma}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"notallowed": {"message": "

Input as a fraction or an integer and not as a decimal

", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{2*b*c}/{2*b-a}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "\n

Let $I=[\\var{l},\\var{m}]$ be an interval and let $g: I \\rightarrow \\mathbb{R}$ be the function given by:
\\[g(x)=\\simplify{x^2/(x-{c})^({a}/{b})}\\]

\n

Answer the following questions. There are seven parts and you may need to scroll down to complete all parts.

\n

 

\n ", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "if(b=3,random(1,2,4,5),if(b=5,random(1,2,3,4,6,7,8,9),random(2,3,4,5,6,8,9)))", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "round(2b/a)+random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "random(3,5,7)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "valmin": {"definition": "precround(statpoint^2/(statpoint-c)^(a/b),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valmin", "description": ""}, "m": {"definition": "round(statpoint)+random(2..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "m", "description": ""}, "l": {"definition": "if(l1=c,l1+1,l1)", "templateType": "anything", "group": "Ungrouped variables", "name": "l", "description": ""}, "xma": {"definition": "if(valend>valbegin,m,l)", "templateType": "anything", "group": "Ungrouped variables", "name": "xma", "description": ""}, "lmi": {"definition": "if(a<0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lmi", "description": ""}, "s": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s", "description": ""}, "valend": {"definition": "precround(m^2/(m-c)^(a/b),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valend", "description": ""}, "u": {"definition": "random(0..90)", "templateType": "anything", "group": "Ungrouped variables", "name": "u", "description": ""}, "valsd": {"definition": "precround(2*(2*b/a-1)^(a/b+2)/c^(a/b),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valsd", "description": ""}, "tol": {"definition": "0.001", "templateType": "anything", "group": "Ungrouped variables", "name": "tol", "description": ""}, "l1": {"definition": "round((u*(c+1)+(100-u)*round(statpoint-1))/100)", "templateType": "anything", "group": "Ungrouped variables", "name": "l1", "description": ""}, "valbegin": {"definition": "precround(l^2/(l-c)^(a/b),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "valbegin", "description": ""}, "lma": {"definition": "if(a>0,b,-b)", "templateType": "anything", "group": "Ungrouped variables", "name": "lma", "description": ""}, "xmi": {"definition": "lmi", "templateType": "anything", "group": "Ungrouped variables", "name": "xmi", "description": ""}, "statpoint": {"definition": "2*b*c/(2*b-a)", "templateType": "anything", "group": "Ungrouped variables", "name": "statpoint", "description": ""}}, "metadata": {"notes": "\n \t\t

9/07/2012:

\n \t\t

Added tags.

\n \t\t

Corrected mistake in definition of variable valsd. Changed the number of decimal places to 5 for this variable as can be very small and positive.

\n \t\t

Modified display in Advice slightly.

\n \t\t

Set new variable tolerance to be tol=0.001 for entries to 3 dps.

\n \t\t

 

\n \t\t

10/07/2012:

\n \t\t

Added tags.

\n \t\t

In Advice section, increased size of brackets so that they were big enough to contain a fraction.

\n \t\t

Question appears to be working correctly.

\n \t\t

 

\n \t\t", "description": "

$I$ compact interval. $\\displaystyle g: I\\rightarrow I, g(x)=\\frac{x^2}{(x-c)^{a/b}}$. Are there stationary points and local maxima, minima? Has $g$ a global max, global min? 

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "extensions": [], "custom_part_types": [], "resources": []}