Question

In Problems $7-44$, follow Steps 1 through 7 on page 228 to analyze the graph of each function.

$f\left(x\right)=x+\frac{1}{x^3}$

Short answer

The graph is

Step-by-step solution

Given information

Given function$f\left(x\right)=x+\frac{1}{x^3}$

Factoring the numerator and denominator

Factoring, we get

$$\begin{gathered} f\left(x\right)=x+\frac{1}{x^3} \\ f\left(x\right)=\frac{x^4+1}{x^3} \end{gathered}$$

Calculate the domain

Setting the denominator to zero, we get

$$\begin{gathered} x^3=0 \\ x=0 \end{gathered}$$

The domain is all real numbers except$0$

Calculate the x intercept

Setting numerator to zero, we get

$$\begin{gathered} x^4+1=0 \\ {x}^4=-1 \end{gathered}$$

There is no$x,y$intercept

Calculate the vertical asymptote

The zeroes of denominator is $x=0$so the vertical asymptote is$x=0$

Calculating the horizontal asymptote

The degree of numerator is greater than denominator so there is an oblique asymptote

$f\left(x\right)=x+\frac{1}{x^3}$

The oblique asymptote is $y=x$

$$\begin{gathered} f\left(x\right)=\frac{\left(x^4+1\right)}{x} \\ =x \\ {x}^4+1=x^2 \\ {x}^4-x^2=-1 \end{gathered}$$

It doesn't intersects$f$

Sketching the graph

The graph is