Chapter 4.3, Problem 44AYU

To determine

To analyze: the graph of given function

Answer to Problem 44AYU

  

Explanation of Solution

Given:

  $f\left(x\right)=2x+\frac{9}{x^3}$

Calculation:

STEP 1: Rewrite the function as $f\left(x\right)=\frac{x^4+1}{x^3}$ . We can see that $f$ is completely factored.

The domain of $f$ is $\{x|x\cancel{=}0\}$ .

Since $0$ is not the domain, there is no $y-intercept$ .

STEP 2: The function $f$ is in lowest terms. Since $x^4+1=0$ has no real solution, there are no $x-\mathrm{intercepts}$ .

STEP 3: $f$ is in lowest terms. So, $x=0$ is a vertical asymptote of $f$ . Graph the line $x=0$ using a dashed line.

STEP 4: Degree of the numerator is greater than the degree of the denominator. So, the rational function is improper. To find any horizontal or oblique asymptote, we use long division.

  $x^3\begin{array}{c}\hfill x\\ \hfill \overline{)\begin{array}{l}{x}^4+1\\ \frac{x^4}{\text{ 1}}\end{array}}\end{array}$

The quotient is $x$ . So, the graph has theoblique asymptote $y=x$ . But the graph of $f$ will approach the graph of $y=x\text{ as }x\to -\infty andx\to \infty$ . The graph of $f$ does not intersect.Graph $y=x$ Using a dashed line.

STEP 5: The numerator has no zero and the denominator has one zero at $0$ . Therefore, we divide the $x-axis$ into two intervals. The two intervals are $\left(-\infty ,0\right),\left(0,\infty \right)$ .

Now, construct the table.

  

Interval Number chosen Value of f Location of graph Point on graph

$\begin{array}{l}\left(-\infty ,0\right)\\ -1\\ f\left(-1\right)=-2\\ \text{Below }x-axis\\ \left(-1,-2\right)\end{array}$

$\begin{array}{l}\left(0,\infty \right)\\ 1\\ f\left(1\right)=2\\ \text{Above }x-axis\\ \left(1,2\right)\end{array}$

Plot the points from the table obtained.

  

STEP 6: Since the graph of $f$ is below the $x-axisforx<0$ and does not intersect $y=x$ , we place arrows below $y=x$ . Similarly, as the graph of $f$ is above the $x-axisforx>0$ , anddoes not intersect $y=x$ we place arrows above

  $y=x$ .

Also, as the graph will approach the vertical asymptote $x=0$ at the bottom to the left of $x=0$ , and to the right of $x=0$ at the top.

  

STEP 7: The figure shows the final graph.

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 6.