# 009A Sample Final 2, Problem 10

Let

${\displaystyle f(x)={\frac {4x}{x^{2}+1}}}$

(a) Find all local maximum and local minimum values of  ${\displaystyle f,}$  find all intervals where  ${\displaystyle f}$  is increasing and all intervals where  ${\displaystyle f}$  is decreasing.

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

(c) Find all horizontal asymptotes of the graph  ${\displaystyle y=f(x).}$

(d) Sketch the graph of  ${\displaystyle y=f(x).}$

Foundations:
1. ${\displaystyle f(x)}$  is increasing when  ${\displaystyle f'(x)>0}$  and  ${\displaystyle f(x)}$  is decreasing when  ${\displaystyle f'(x)<0.}$
2. The First Derivative Test tells us when we have a local maximum or local minimum.
3. ${\displaystyle f(x)}$  is concave up when  ${\displaystyle f''(x)>0}$  and  ${\displaystyle f(x)}$  is concave down when  ${\displaystyle f''(x)<0.}$
4. Inflection points occur when  ${\displaystyle f''(x)=0.}$

Solution:

(a)

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

(b)

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

(c)

Step 1:
By L'Hopital's Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }f(x)}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {4x}{x^{2}+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {4}{2x}}}\\&&\\&=&\displaystyle {0.}\end{array}}}$
Similarly, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }f(x)}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {4x}{x^{2}+1}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {4}{2x}}}\\&&\\&=&\displaystyle {0.}\end{array}}}$
Step 2:
Since
${\displaystyle \displaystyle {\lim _{x\rightarrow -\infty }f(x)=\lim _{x\rightarrow \infty }f(x)=0,}}$
${\displaystyle f(x)}$  has a horizontal asymptote
${\displaystyle y=0.}$
(d):

(a)    ${\displaystyle f(x)}$  is increasing on  ${\displaystyle (-1,1)}$  and decreasing on  ${\displaystyle (-\infty ,-1)\cup (1,\infty ).}$
The local maximum value of  ${\displaystyle f}$  is  ${\displaystyle 2}$  and the local minimum value of  ${\displaystyle f}$  is  ${\displaystyle -2}$
(b)    ${\displaystyle f(x)}$  is concave up on  ${\displaystyle (-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty ),}$  and concave down on  ${\displaystyle (-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}}).}$
The inflection points are  ${\displaystyle (0,0),(-{\sqrt {3}},-{\sqrt {3}}),({\sqrt {3}},{\sqrt {3}}).}$
(c)    ${\displaystyle y=0}$