Q. 70 - Calculus

Question

Sketch careful, labeled graphs of each function $f$ in Exercises $57-82$ by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f$ and $f^{\text{'}}$and examine any relevant limits so that you can describe all key points and behaviors of $f$.

$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$.

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$ is,

Step 1 . Given information

$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$.

Step 2 . Let y = 1 x 2 - 1 .

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-2$	$\frac{1}{\sqrt{8}}$	$\left(-2,\frac{1}{8}\right)$
$-3$	$\frac{1}{\sqrt{3}}$	$\left(-3,\frac{1}{\sqrt{3}}\right)$
$2$	$\frac{1}{\sqrt{3}}$	$\left(2,\frac{1}{\sqrt{3}}\right)$
$3$	$\frac{1}{\sqrt{8}}$	$\left(3,\frac{1}{\sqrt{8}}\right)$

Step 3 . The graph for the function is,

Step 4 . Now for critical point f ' x = 0 .

$$\begin{gathered} \frac{d}{dx}\left(\frac{1}{\sqrt{x^2-1}}\right)=0 \\ -\frac{1}{2}{\left(x^2-1\right)}^{\frac{3}{2}}.2x=0 \\ -\frac{x}{{\left(x^2-1\right)}^{{\displaystyle \frac{3}{2}}}}=0 \\ x=0 \end{gathered}$$

Therefore, $f$ has a critical point at $x=0$.So, it has no local extrema.

Step 5 . The sign chart of f is shown below:

For roots of the function,

$\frac{1}{\sqrt{x^2-1}}=0$

Therefore, $x$ has no real value.

Step 6 . Therefore, the function f is increasing on - ∞ , 1 and decreasing on 1 , ∞ .

Again,

$$\begin{gathered} \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\frac{1}{\sqrt{x^2-1}} \\ =0 \\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{1}{\sqrt{x^2-1}} \\ =0 \end{gathered}$$

Therefore, the function is defined only on $\left(-\infty ,-1\right)\cup \left(1,\infty \right)$ and undefined on $\left(-1,1\right)$.It is positive everywhere.