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$g: \\mathbb{R} \\rightarrow \\mathbb{R}, g(x)=\\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min?
"}, "statement": "Let $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ be the function given by:
\\[g(x)=\\simplify{{a}*x/(x^2+{b}^2)}\\]
The function $g(x)$ is continuous and differentiable at all points in $\\mathbb{R}$.
\nUsing 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*} \\]
The stationary points are given by solving $g'(x)=0$.
\n$g'(x)=0 \\Rightarrow \\simplify{{-a}*(x-{b})(x+{b})=0} \\Rightarrow x=\\var{b} \\mbox{ or } x=\\var{-b}$
\nThe second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:
\nUsing 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*} \\]
The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.
\nFor $x= \\var{lma}$ we have: \\[g''(\\var{lma})= \\simplify[std]{-{abs(a)}/{2*b^3}} \\lt 0\\]
\nHence is a local maximum.
\nEvaluating the function at $x=\\var{lma}$ gives $g(\\var{lma})=\\var{valmax}$ to 3 decimal places.
\nFor $x= \\var{lmi}$ we have: \\[g''(\\var{lmi})= \\simplify[std]{{abs(a)}/{2*b^3}} \\gt 0\\]
\nHence is a local minimum.
\nEvaluating the function at $x=\\var{lmi}$ gives $g(\\var{lmi})=\\var{valmin}$ to 3 decimal places.
\nIf 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)}\\]
\nThen 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$\\lim_{x \\to -\\infty}g(x)=0$
\nSince $g$ 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 $\\var{valmax}$
\nThen:
\nGlobal Maximum: The local maximum of $g$ we have found at $x=\\var{lma}$ must be a global maximum and similarly,
\nGlobal Minimum: The local minimum of $g$ we have found at $x=\\var{lmi}$ must be a global minimum.
\nSo we have shown \\[\\forall x \\in \\mathbb{R},\\;\\;\\var{valmin} \\le g(x) \\le \\var{valmax}\\]
\n ", "name": "Paul 's copy of Max and Min 4", "parts": [{"matrix": [1, 0], "showCorrectAnswer": true, "marks": 0, "useCustomName": false, "choices": ["Yes
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\n \nChoose Yes or No.
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\nInput 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 $c$:
$p(x)\\;=\\;$[[0]]
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\n \nChoose Yes or No.
\n \n ", "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "shuffleChoices": true, "extendBaseMarkingAlgorithm": true, "maxMarks": 0, "customMarkingAlgorithm": "", "unitTests": []}, {"gaps": [{"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "valuegenerators": [], "useCustomName": false, "scripts": {}, "customName": "", "type": "jme", "checkingType": "absdiff", "variableReplacementStrategy": "originalfirst", "vsetRangePoints": 5, "showFeedbackIcon": true, "checkVariableNames": false, "vsetRange": [0, 1], "showPreview": true, "checkingAccuracy": 0.001, "answerSimplification": "std", "failureRate": 1, "answer": "{-b}", "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": []}, {"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "valuegenerators": [], "useCustomName": false, "scripts": {}, "customName": "", "type": "jme", "checkingType": "absdiff", "variableReplacementStrategy": "originalfirst", "vsetRangePoints": 5, "showFeedbackIcon": true, "checkVariableNames": false, "vsetRange": [0, 1], "showPreview": true, "checkingAccuracy": 0.001, "answerSimplification": "std", "failureRate": 1, "answer": "{b}", "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": []}], "sortAnswers": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "prompt": "\nLeast stationary point: [[0]]
\n \nGreatest stationary point: [[1]]
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\nInput 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 $a_1$, $a_2$
$r(x)=\\;\\;$ [[0]]
\nHence find all local maxima and minima given by the stationary points
\nLocal maximum is at $x=\\;\\;$ [[1]] and the value of the function at the local maximum (to 3 decimal places)= [[2]]
\nLocal minimum is at $x=\\;\\;$ [[3]] and the value of the function at the local minimum (to 3 decimal places) = [[4]]
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\n1) $\\lim_{x \\to \\infty}g(x)\\;\\;$
\nChoose one of the following [[0]]
\n2) $\\lim_{x \\to -\\infty}g(x)$
\nChoose one of the following [[1]]
\nDoes $g$ have a finite global maximum? Click on Yes or No
[[2]]
Does $g$ have a finite global maximum? Click on Yes or No
[[2]]