Chapter 2.8, Problem 28E

Solution

(a).

To find: The real values of $x$ that cause the function to be zero.

The real values of $x$ are equal to $0,-\frac{9}{2}$ .

Given information:

Consider the given function.

  $f\left(x\right)=x^2\left|2x+9\right|$

Calculation:

When the given function $f\left(x\right)$ is zero, then the values of $x$ can be calculated below:

  $\begin{array}{c}{x}^2\left|2x+9\right|=0\\ x^2=0\\ x=0\end{array}$

And

  $\begin{array}{c}\left|2x+9\right|=0\\ 2x+9=0\\ 2x=-9\\ x=-\frac{9}{2}\end{array}$

Thus, the real values of $x$ are equal to $0,-\frac{9}{2}$ .

(b).

To find: The real values of $x$ that cause the function to be undefined.

No solution for the function $f\left(x\right)$ is undefined.

Given information:

Consider the given function.

  $f\left(x\right)=x^2\left|2x+9\right|$

Calculation:

When the given function $f\left(x\right)$ is undefined, then the values of $x$ can be calculated below:

  $\begin{array}{c}{x}^2\left|2x+9\right|=\frac{1}{0}\\ \text{No}\text{solution}\end{array}$

Therefore, no solution for the function $f\left(x\right)$ is undefined.

(c).

To find: The real values of $x$ that cause the function to be positive.

The real values of $x$ is equal to $\left(-\infty ,-\frac{9}{2}\right)\cup \left(-\frac{9}{2},0\right)\cup \left(0,\infty \right)$ .

Given information:

Consider the given function.

  $f\left(x\right)=x^2\left|2x+9\right|$

Calculation:

The given function $f\left(x\right)$ is positive.

First find the zeros of the function $f\left(x\right)$ .

Substitute $f\left(x\right)=0$ and calculate the values of $x$ :

  $\begin{array}{c}{x}^2\left|2x+9\right|=0\\ x^2=0\\ x=0\end{array}$

And

  $\begin{array}{c}\left|2x+9\right|=0\\ 2x+9=0\\ 2x=-9\\ x=-\frac{9}{2}\end{array}$

So, the zeros of the function $f\left(x\right)$ are $0,-\frac{9}{2}$ .

Now, locate these zeros on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

Therefore, the given function $f\left(x\right)$ is positive into $\left(-\infty ,-\frac{9}{2}\right)\cup \left(-\frac{9}{2},0\right)\cup \left(0,\infty \right)$ .

Hence, the real values of $x$ is equal to $\left(-\infty ,-\frac{9}{2}\right)\cup \left(-\frac{9}{2},0\right)\cup \left(0,\infty \right)$ .

(d).

To find: The real values of $x$ that cause the function to be negative.

There are no real values of $x$ .

Given information:

Consider the given function.

  $f\left(x\right)=x\sqrt{x+3}$

Calculation:

The given function $f\left(x\right)$ is negative.

First find the zeros of the function $f\left(x\right)$ .

Substitute $f\left(x\right)=0$ and calculate the values of $x$ :

  $\begin{array}{c}{x}^2\left|2x+9\right|=0\\ x^2=0\\ x=0\end{array}$

And

  $\begin{array}{c}\left|2x+9\right|=0\\ 2x+9=0\\ 2x=-9\\ x=-\frac{9}{2}\end{array}$

So, the zeros of the function $f\left(x\right)$ are $0,-\frac{9}{2}$ .

Now, locate these zeros on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

Therefore, there is no value of $x$ , where the given function $f\left(x\right)$ is negative.

Hence, there are no real values of $x$ .