Chapter 6.5, Problem 49AYU

In Problems 49 and 50, graph each function.

$f\left(x\right)=\{\begin{array}{l}\tan x\text{0}\le x\text{<}\frac{\pi }{2}\\ 0x\text{=}\frac{\pi }{2}\\ \sec x\frac{\pi }{2}\text{<}x\le \pi \end{array}$

To determine

To find: Graph the function $f\left(x\right)$.

Answer to Problem 49AYU

Solution:

Explanation of Solution

Given:

$f\left(x\right)=\{\begin{array}{c}\tan x 0\le x\le \frac{\pi }{2}\\ 0 x=\frac{\pi }{2} \\ \sec x \frac{\pi }{2}\le x\le \pi \end{array}$

Calculation:

Here we calculate some points for $\tan x$ and $\sec x$ in given interval, as shown in the table.

$x$	$\tan x$	$(x,\tan x)$
$\frac{\pi }{4}$	$1$	$(\frac{\pi }{4},1)$
$\frac{\pi }{3}$	$\sqrt{3}$	$(\frac{\pi }{3},\sqrt{3})$
$0$	$0$	$\left(0,0\right)$

$x$	$\sec x$	$(x,\sec x)$
$\frac{3\pi }{4}$	$-\sqrt{2}$	$(\frac{3\pi }{4},-\sqrt{2})$
$\frac{5\pi }{6}$	$-\frac{2}{\sqrt{3}}$	$(\frac{5\pi }{6},-\frac{2}{\sqrt{3}})$
$\pi$	$-1$	$(\pi ,-1)$

Now plotting above points on the graph, applying other features of tangent and secant functions, the required graph looks as shown below.