# 009A Sample Final 2, Problem 10

Let

$f(x)={\frac {4x}{x^{2}+1}}$ (a) Find all local maximum and local minimum values of  $f,$ find all intervals where  $f$ is increasing and all intervals where  $f$ is decreasing.

(b) Find all inflection points of the function  $f,$ find all intervals where the function  $f$ is concave upward and all intervals where  $f$ is concave downward.

(c) Find all horizontal asymptotes of the graph  $y=f(x).$ (d) Sketch the graph of  $y=f(x).$ Foundations:
1. $f(x)$ is increasing when  $f'(x)>0$ and  $f(x)$ is decreasing when  $f'(x)<0.$ 2. The First Derivative Test tells us when we have a local maximum or local minimum.
3. $f(x)$ is concave up when  $f''(x)>0$ and  $f(x)$ is concave down when  $f''(x)<0.$ 4. Inflection points occur when  $f''(x)=0.$ Solution:

(a)

Step 1:
We start by taking the derivative of  $f(x).$ Using the Quotient Rule, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {(x^{2}+1)(4x)'-(4x)(x^{2}+1)'}{(x^{2}+1)^{2}}}\\&&\\&=&\displaystyle {\frac {(x^{2}+1)(4)-(4x)(2x)}{(x^{2}+1)^{2}}}\\&&\\&=&\displaystyle {{\frac {-4(x^{2}-1)}{(x^{2}+1)^{2}}}.}\end{array}}$ Now, we set  $f'(x)=0.$ So, we have
$-4(x^{2}-1)=0.$ Hence, we have  $x=1$ and  $x=-1.$ So, these values of  $x$ break up the number line into 3 intervals:
$(-\infty ,-1),(-1,1),(1,\infty ).$ Step 2:
To check whether the function is increasing or decreasing in these intervals, we use testpoints.
For  $x=-2,~f'(x)={\frac {-12}{25}}<0.$ For  $x=0,~f'(x)=4>0.$ For  $x=2,~f'(x)={\frac {-12}{25}}<0.$ Thus,  $f(x)$ is increasing on  $(-1,1)$ and decreasing on  $(-\infty ,-1)\cup (1,\infty ).$ Step 3:
Using the First Derivative Test,  $f(x)$ has a local minimum at  $x=-1$ and a local maximum at  $x=1.$ Thus, the local maximum and local minimum values of  $f$ are
$f(1)=2,~f(-1)=-2.$ (b)

Step 1:
To find the intervals when the function is concave up or concave down, we need to find  $f''(x).$ Using the Quotient Rule and Chain Rule, we have
${\begin{array}{rcl}\displaystyle {f''(x)}&=&\displaystyle {\frac {(x^{2}+1)^{2}(-4(x^{2}-1))'+4(x^{2}-1)((x^{2}+1)^{2})'}{((x^{2}+1)^{2})^{2}}}\\&&\\&=&\displaystyle {\frac {(x^{2}+1)^{2}(-8x)+4(x^{2}-1)2(x^{2}+1)(x^{2}+1)'}{(x^{2}+1)^{4}}}\\&&\\&=&\displaystyle {\frac {(x^{2}+1)^{2}(-8x)+4(x^{2}-1)2(x^{2}+1)(2x)}{(x^{2}+1)^{4}}}\\&&\\&=&\displaystyle {\frac {(x^{2}+1)(-8x)+16(x^{2}-1)x}{(x^{2}+1)^{3}}}\\&&\\&=&\displaystyle {\frac {8x^{3}-24x}{(x^{2}+1)^{3}}}\\&&\\&=&\displaystyle {{\frac {8x(x^{2}-3)}{(x^{2}+1)^{3}}}.}\end{array}}$ We set  $f''(x)=0.$ So, we have
$8x(x^{2}-3)=0.$ Hence,
$x=0,~x=-{\sqrt {3}},~x={\sqrt {3}}.$ This value breaks up the number line into four intervals:
$(-\infty ,-{\sqrt {3}}),(-{\sqrt {3}},0),(0,{\sqrt {3}}),({\sqrt {3}},\infty ).$ Step 2:
Again, we use test points in these four intervals.
For  $x=-2,$ we have  $f''(x)={\frac {-16}{125}}<0.$ For  $x=-1,$ we have  $f''(x)=2>0.$ For  $x=1,$ we have  $f''(x)=-2<0.$ For  $x=2,$ we have  $f''(x)={\frac {16}{125}}>0.$ Thus,  $f(x)$ is concave up on  $(-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty ),$ and concave down on  $(-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}}).$ Step 3:
The inflection points occur at
$x=0,~x=-{\sqrt {3}},~x={\sqrt {3}}.$ Plugging these into  $f(x),$ we get the inflection points
$(0,0),(-{\sqrt {3}},-{\sqrt {3}}),({\sqrt {3}},{\sqrt {3}}).$ (c)

Step 1:
By L'Hopital's Rule, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }f(x)}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {4x}{x^{2}+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {4}{2x}}}\\&&\\&=&\displaystyle {0.}\end{array}}$ Similarly, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }f(x)}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {4x}{x^{2}+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {4}{2x}}}\\&&\\&=&\displaystyle {0.}\end{array}}$ Step 2:
Since
$\displaystyle {\lim _{x\rightarrow -\infty }f(x)=\lim _{x\rightarrow \infty }f(x)=0,}$ $f(x)$ has a horizontal asymptote
$y=0.$ (a)    $f(x)$ is increasing on  $(-1,1)$ and decreasing on  $(-\infty ,-1)\cup (1,\infty ).$ The local maximum value of  $f$ is  $2$ and the local minimum value of  $f$ is  $-2$ (b)    $f(x)$ is concave up on  $(-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty ),$ and concave down on  $(-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}}).$ The inflection points are  $(0,0),(-{\sqrt {3}},-{\sqrt {3}}),({\sqrt {3}},{\sqrt {3}}).$ (c)    $y=0$ 