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
$y=0.$

(d):
Insert sketch

Final Answer:
(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)    $y=0$
(d)    See above

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