// Numbas version: exam_results_page_options {"name": "CF Maths January test mock paper finding stationary points", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "c", "d", "gma", "gmi", "l", "l1", "m", "m1", "rawvalbegin", "rawvalend", "rawvalmax", "rawvalmin", "rtemp1", "rtemp2", "s", "temp1", "temp2", "valbegin", "valend", "valgmax", "valgmin", "valmax", "valmin"], "name": "CF Maths January test mock paper finding stationary points", "tags": ["Calculus", "calculus", "classifying stationary points", "finding global maxima and minima", "finding local maxima and local minima", "finding stationary points", "finding the maximum and minimum of a function", "nature of a critical point", "nature of critical points", "nature of turning points", "optimisation", "optimising a function on an interval", "optimising functions", "stationary points", "turning points"], "advice": "

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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Differentiate the function $g(x)$.

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Input the first derivative of $g$ here,

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 $g'(x)=\\;\\;$[[0]]

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Find the stationary points of $g$ .

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the lower value of $x$ at the stationary point is:  [[0]] 

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the bigger value of $x$ at the stationary point is: [[1]]

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Input the second derivative of $g$:

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

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Local maximum is at $x=\\;\\;$ [[1]] ,

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and the value of the function ($g(x)$)at the local maximum:

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$g(x)=\\;\\;$ [[2]]

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Local minimum is at $x=\\;\\;$ [[3]] ,

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and the value of the function ($g(x)$) at the local minimum:

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$g(x)=\\;\\;$ [[4]]

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Find and indentify turning points of the function given by :

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\\[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.

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Question appears to be working correctly.

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Changed grammar in the Advice section.

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$I$ 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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