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  $f''(x).$
We have  $f''(x)=6x-12.$
We set  $f''(x)=0.$
So, we have  $0=6x-12.$  Hence,  $x=2.$
This value breaks up the number line into two intervals:  $(-\infty ,2),(2,\infty ).$

Step 2:

(c)

Step 1:
First, we note that the degree of the numerator is  $1$  and
the degree of the denominator is  $2.$

Step 2:
Since the degree of the denominator is greater than the degree of the numerator,
$f(x)$  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)
(c)    $y=0$
(d)    See above

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