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

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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*} \\]

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Stationary Points.

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

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

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

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

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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*} \\]

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

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

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Hence is a local maximum.

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

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

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Hence is a local minimum.

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

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The values of $g$ at the endpoints are:

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

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Global Maximum and Minimum

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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 $I$).

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then:

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

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

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

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(all to 3 decimal places).

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

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Choose Yes or No.

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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.

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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 $c$:

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

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Factorise the expression

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Factorise the expression

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

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Choose Yes or No.

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

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Find the stationary points of $g$.

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Least stationary point: [[0]]

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Greatest stationary point: [[1]]

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

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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.

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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 $a_1$, $a_2$

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$r(x)=\\;\\;$ [[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 (to 3 decimal places)= [[2]]

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

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Factorise the expression

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Factorise the expression

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

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

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

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Input both to 3 decimal places.

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

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

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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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Global Minimum

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

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

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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)}\\]

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

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9/07/2012:

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Added tags.

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

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

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10/07/2012:

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Added tags.

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Edited grammar in the Advice section.

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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? 

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