Q. 66 - 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 $f$.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . Let y = x - 1 x .

Now point the table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$1$	$0$	$\left(1,0\right)$
$2$	$0.4142$	$\left(2,0.4142\right)$
$4$	$1.5$	$\left(4,1.5\right)$

Step 3 . The graph for the function is,

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

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

Therefore, $f$ has a critical point at $x=0$. But that is a local minimum at $x=0$.

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

For roots of the function,

$$\begin{gathered} \sqrt{x}-\frac{1}{\sqrt{x}}=0 \\ \frac{x-1}{\sqrt{x}}=0 \\ x-1=0 \\ x=1 \end{gathered}$$

Step 6 . Therefore, the function f is increasing on - ∞ , 0 and decreasing on 0 , ∞ .

Again,

$\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$

                   $=$Doesn't exist.

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

Therefore, the function is defined on $\left(0,\infty \right)$ and the root at $x=-1$. Positive on $\left(1,\infty \right)$ and doesn't exist at $\left(-\infty ,0\right)$.