Question

Use the first-derivative test to determine the local extrema of each function $f$in Exercises $39-50$. Then verify your algebraic answers with graphs from a calculator or graphing utility.

$f\left(x\right)=arc\tan x$.

Short answer

The function $f\left(x\right)=arc\tan x$has no local extrema. The following graph verifies the algebraic result graphically:

Step-by-step solution

Step 1. Given information

$f\left(x\right)=arc\tan x$.

Step 2. Consider the function,

$f\left(x\right)=arc\tan x$.

First find the derivative for the given function.

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(arc\tan x\right) \\ =\frac{1}{1+x^2} \end{gathered}$$

The derivative is defined and continuous everywhere, so the critical points of $f$are just the points where role="math" localid="1648397112503" $f^{\text{'}}\left(x\right)=0$that is,

$f^{\text{'}}\left(x\right)=0$

Therefore, there is no critical point for which $f^{\text{'}}\left(x\right)=0$.

Hence, there is no local extrema.

Step 3. The following graph verifies the algebraic result graphically: