// Numbas version: finer_feedback_settings {"name": "Max and Min (6 parts) 24/11", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "gma", "k", "m", "xma", "l", "valend", "gmi", "valbegin", "third2", "third1", "s1"], "name": "Max and Min (6 parts) 24/11", "tags": ["Calculus", "calculus", "classifying stationary points", "finding global maxima and minima", "finding local mamima and minima", "finding the stationary points", "optimisation", "optimising functions", "third derivative test for maximum or minimum"], "advice": "

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.

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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 $I$.

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Second derivative.

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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 $g(\\var{b})= 0$.

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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

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

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More information on maxima and minima is chapter 3 of your Mathematical Physics book, sections 3.2 and 3.3, pages 51-53.

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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 $c$ (these can be zero):

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

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Factorise the expression

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Factorise the expression

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Find the stationary points of $g$,

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Least stationary point is at $x$ =[[0]]

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Greatest stationary point is at $x$ = [[1]]

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

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Yes

", "

No

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Input the second derivative of $g$:

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

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

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

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

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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]]

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

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

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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.

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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 ", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "variableReplacements": [], "maxValue": "{valbegin}", "minValue": "{valbegin}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 0.5, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{valend}", "minValue": "{valend}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 0.5, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "

Global Maximum

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

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

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

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Global Minimum

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

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

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

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

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9/07/2012:

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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 $I$. 

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