Q. 68 - 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 $f$.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

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

$y=\frac{x+1}{x^2}$.

Now point the table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-2$	$-\frac{1}{4}$	$\left(-2,-\frac{1}{4}\right)$
$-1$	$0$	$\left(-1,0\right)$
$1$	$2$	$\left(1,2\right)$
$2$	$\frac{3}{4}$	$\left(2,\frac{3}{4}\right)$
$3$	$\frac{4}{9}$	$\left(3,\frac{4}{9}\right)$

Step 3 . The graph for the function is,

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

$$\begin{gathered} \frac{d}{dx}\left(\frac{x+1}{x^2}\right)=0 \\ \frac{x^2\left(x+1\right)-\left(x+1\right)\left(2x\right)}{{\left(x^2\right)}^2}=0 \\ \frac{x^3+x^2-2x^2-2x}{x^4}=0 \\ \frac{x^3-x^2-2x}{x^4}=0 \\ \frac{x^2-x-2}{x^3}=0 \\ {x}^2-x-2=0 \\ {x}^2-2x+x-2=0 \\ x\left(x-2\right)+1\left(x-2\right)=0 \\ \left(x+1\right)\left(x-2\right)=0 \\ x=2,-1 \end{gathered}$$

Therefore, $f$ has two critical points. So, it has no local extrema.

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

For roots of the function,

$$\begin{gathered} \frac{x+1}{x^2}=0 \\ x+1=0 \\ x=-1 \end{gathered}$$

Step 6 . Therefore, the function f is increasing on - 1 , 0 and decreasing on - ∞ , - 1 ∪ 0 , ∞ .

Again,

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

Therefore, the function $f$ is defined everywhere except at $x=0$ and the root at $x=-1$.Positive on $\left(-1,\infty \right)$ and negative on $\left(-\infty ,-1\right)$.

The limits are, $\underset{x\to -\infty }{\lim }f\left(x\right)=0 ;\underset{x\to \infty }{\lim }f\left(x\right)=0$. So, there is a horizontal asymptote at $y=0$.