// Numbas version: finer_feedback_settings {"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:
\n\\[g'(x)=\\simplify{{c}*x^2+{-c*(a+b)}*x+{c*a*b}={c}*(x+{-a})(x-{b})}\\]
\nNote that we have already factorised the derivative.
\nStationary points are given by solving $g'(x)=0 \\Rightarrow x=\\var{a},\\;\\;\\mbox{or }x=\\var{b}$
\nSo the least stationary point is $x=\\var{a}$ and the greatest is $x=\\var{b}$.
\nSince $\\var{a} > \\var{l}$ and $\\var{b} \\lt \\var{m}$ we have that both stationary points are in $I$.
\nThe second derivative is given by \\[g''(x)=\\simplify{{2*c}*x-{c*(a+b)}}\\]
\nAt the stationary point $x=\\var{a}$ we have $g''(\\var{a})=\\var{c*a-c*b} \\lt 0$.
\nHence at this value of $x$ we have a local maximum.
\nThe value of the function $g$ at this local maximum is $g(\\var{a})= \\var{valmax}$.
\nAt the stationary point $x=\\var{b}$ we have $g''(\\var{b})=\\var{c*b-c*a} \\gt 0$.
\nHence this point is a local minimum.
\nThe value of the function $g$ at this local minimum is $g(\\var{b})= \\var{valmin}$.
\nFirst we find the values at the endpoints of the interval $I=[\\var{l},\\var{m}]$ are:
\n$g(\\var{l})=\\var{valbegin}$ to 3 decimal places.
\n$g(\\var{m})=\\var{valend}$ to 3 decimal places.
\nTo find the global maximum note that we are only concerned with the values of $g$ on the interval $I$.
\nSo we proceed by comparing the values of the function at the endpoints with the local maximum.
\na) 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.
\nb) Otherwise if the greatest value of the function at the endpoints is greater than the local maximum then this is the global maximum.
\n\\[\\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} \\]
\nSo for our example we see that the global maximum occurs at $x=\\var{gma}$ and $g(\\var{gma})=\\var{valgmax}$.
\nWe proceed as for the global maximum by comparing the values of the function at the endpoints with the local minimum.
\na) 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.
\nb) Otherwise if the least value of the function at the endpoints is less than the local minimum then this is the global minimum.
\n\\[\\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} \\]
\nIn our example we see that the global minimum occurs at $x=\\var{gmi}$ and $g(\\var{gmi})=\\var{valgmin}$.
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\n$g'(x)=\\;\\;$[[0]]
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\nthe bigger value of $x$ at the stationary point is: [[1]]
\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"vsetrangepoints": 5, "expectedvariablenames": [], "checkingaccuracy": 0.001, "type": "jme", "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "vsetrange": [0, 1]}, {"vsetrangepoints": 5, "expectedvariablenames": [], "checkingaccuracy": 0.001, "type": "jme", "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{b}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "vsetrange": [0, 1]}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "Input the second derivative of $g$:
\n$g''(x)=\\;\\;$ [[0]]
\nLocal maximum is at $x=\\;\\;$ [[1]] ,
\nand the value of the function ($g(x)$)at the local maximum:
\n$g(x)=\\;\\;$ [[2]]
\nLocal minimum is at $x=\\;\\;$ [[3]] ,
\nand the value of the function ($g(x)$) at the local minimum:
\n$g(x)=\\;\\;$ [[4]]
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\n\\[g(x) = \\simplify{{c}/3*x^3+ {-c*(a+b)}/2*x^2+{c*a*b}*x+{d}}\\]
\n", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"temp2": {"definition": "precround(rtemp2,2)", "templateType": "anything", "group": "Ungrouped variables", "name": "temp2", "description": ""}, "temp1": {"definition": "precround(rtemp1,2)", "templateType": "anything", "group": "Ungrouped variables", "name": "temp1", "description": ""}, "gmi": {"definition": "if(valbegin9/07/2102:
\n \t\tAdded tags.
\n \t\tQuestion appears to be working correctly.
\n \t\tChanged 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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