Question

Intervals of behavior: For each of the following functions $f$, determine the intervals on which$f$ is positive, negative, increasing, decreasing, concave up, and concave down.

$f\left(x\right)={\sec }^{2}x$

Short answer

Therefore, $f\text{'}$is positive on $\dots \cup \left(-\pi ,-\frac{\pi }{2}\right)\cup \left(0,\frac{\pi }{2}\right)\cup \dots$that is $f$is always increasing on $\dots \cup (-\pi ,-\frac{\pi }{2})\cup (0,\frac{\pi }{2})\cup \dots$. And $f\text{'}$is negative on $\dots \cup (-\frac{3\pi }{2},-\pi )\cup (-\frac{\pi }{2},0)\cup \dots$, that is $f$ is always decreasing $\dots \cup (-\frac{3\pi }{2},-\pi )\cup (-\frac{\pi }{2},0)\cup \dots$ The function is always concave up where it is defined.

Step-by-step solution

Step 1. Given data

We have been given the function,

$f\left(x\right)={\sec }^{2}x$

Step 2. Critical points   

We have to find the derivative of the given function$f\left(x\right)={\sec }^{2}x$

Therefore,

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =\frac{d}{dx}({\sec }^{2}x)\\ & =2\sec x\sec x\tan x\\ & =2{\sec }^{2}x\tan x\\ & =2(1+{\tan }^{2}x)\tan x\end{array}$

The derivative is defined and continuous everywhere except at $x=(2k+1)\frac{\pi }{2}$, where $k=0,1,2,3,\dots$.

The function has a critical point at $f^{\text{'}}\left(x\right)=0$.

That is,

$\begin{array}{rl}2(1+{\tan }^{2}x)\tan x& =0\\ (1+{\tan }^{2}x)\tan x& =0\\ \tan x& =0\\ x& =k\pi \end{array}$

These critical points divide the real number line into many parts$(-3\pi ,-2\pi ),(-2\pi ,-\pi )(-\pi ,0),(0,\pi ),(\pi ,2\pi )$

Step 3. Give value for x   

Here, for testing take the sign of $f\text{'}\left(x\right)=0$at one point in that interval.

Let the value of $x$be $x=-\frac{\pi }{4},x=\frac{\pi }{4}$

$\begin{array}{rl}{f}^{\mathrm{\prime }}(-\frac{\pi }{4})& =2(1+{\tan }^2(-\frac{\pi }{4}))\tan (-\frac{\pi }{4})\\ & =2(1+1)(-1)\\ & =-4<0\\ f^{\mathrm{\prime }}\left(\frac{\pi }{4}\right)& =2(1+{\tan }^2\left(\frac{\pi }{4}\right))\tan \left(\frac{\pi }{4}\right)\\ & =2(1+1)\left(1\right)\\ & =4>0\end{array}$

Step 4. Sign chart  

Let us draw the sign chart,

Therefore, $f\text{'}$is positive on $\dots \cup (-\pi ,-\frac{\pi }{2})\cup (0,\frac{\pi }{2})\cup \dots$that is $f$is always increasing on $\dots \cup (-\pi ,-\frac{\pi }{2})\cup (0,\frac{\pi }{2})\cup \dots$. And $f\text{'}$is negative on $\dots \cup (-\frac{3\pi }{2},-\pi )\cup (-\frac{\pi }{2},0)\cup \dots$, that is $f$is always decreasing$\dots \cup (-\frac{3\pi }{2},-\pi )\cup (-\frac{\pi }{2},0)\cup \dots$ The function is always concave up where it is defined.