Chapter 2.4, Problem 47E

To determine

To find: a formula for the function f that satisfies the given conditions.

Answer to Problem 47E

  $f\left(x\right)=-\frac{x-2}{x^2\left(x-3\right)}$

Explanation of Solution

Given:

The given conditions:

  $\begin{array}{l}\underset{x\to \pm \infty }{\lim }f\left(x\right)=0,\underset{x\to 0}{\lim }f\left(x\right)=\pm \infty ,f\left(2\right)=0\\ \underset{x\to 3^{-}}{\lim }f\left(x\right)=\infty ,\underset{x\to 3^{+}}{\lim }f\left(x\right)=-\infty \end{array}$

Calculation:

The given function is $\underset{x\to \pm \infty }{\lim }f\left(x\right)=0$ .

Now, the formula is

  $f\left(x\right)=\frac{1}{x}$

Then the graph is

  

Again, the function is $\underset{x\to 0}{\lim }f\left(x\right)=\pm \infty$

Now, the formula is

  $f\left(x\right)=-\frac{1}{x^2}$

Then the graph is

  

Again, the function is $f\left(2\right)=0$

Now, the formula is

  $f\left(x\right)=\frac{x-2}{x^2}$

Then the graph is

  

Again, the function is $\underset{x\to 3^{-}}{\lim }f\left(x\right)=\infty$

Explanatory Step: Skip if no further instruction needed.

1) Explain Numerators vs. Denominators, and how it affects 0's and asymptotes.

2) Explain why a negative value was dropped from the previous equation.

This step is simply here to explain why I did the things I did. It can be skipped if you are comfortable with graphing asymptotes.

Again, the function is $\underset{x\to 3^{+}}{\lim }f\left(x\right)=-\infty$

Now, the formula is

  $f\left(x\right)=-\frac{x-2}{x^2\left(x-3\right)}$

Then the graph is

  

Therefore, the one possible solution out of the many solution is $f\left(x\right)=-\frac{x-2}{x^2\left(x-3\right)}$ .