Chapter 2.8, Problem 30E

Solution

(a).

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

The real value of $x$ is equal to $1$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{x-1}{\left(x-4\right)\sqrt{x+2}}$

Calculation:

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

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

Thus, the real values of $x$ is equal to $1$ .

(b).

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

The real values of $x$ are equal to $-2,4$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{x-1}{\left(x-4\right)\sqrt{x+2}}$

Calculation:

When the given function $f\left(x\right)$ is undefined, then substitute the denominator equal to zero and calculate the values of $x$ below:

  $\begin{array}{c}\left(x-4\right)\sqrt{x+2}=0\\ x-4=0\\ x=4\end{array}$

And

  $\begin{array}{c}\sqrt{x+2}=0\\ x+2=0\\ x=-2\end{array}$

Therefore, the real values of $x$ are $-2,4$ 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 $\left(-2,1\right)\cup \left(4,\infty \right)$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{x-1}{\left(x-4\right)\sqrt{x+2}}$

Calculation:

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

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

Substitute the numerator and the denominator equal to zero, then calculate the values of $x$ below:

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

And

  $\begin{array}{c}\left(x-4\right)\sqrt{x+2}=0\\ x=-2,4\end{array}$

So, the boundary points of the function $f\left(x\right)$ are $-2,1,4$ .

Now, locate these boundary points 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 $\left(-2,1\right)\cup \left(4,\infty \right)$ .

Hence, the real values of $x$ is equal to $\left(-2,1\right)\cup \left(4,\infty \right)$ .

(d).

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

The real values of $x$ is equal to $\left(1,4\right)$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{x-1}{\left(x-4\right)\sqrt{x+2}}$

Calculation:

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

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

Substitute the numerator and the denominator equal to zero, then calculate the values of $x$ below:

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

And

  $\begin{array}{c}\left(x-4\right)\sqrt{x+2}=0\\ x=-2,4\end{array}$

So, the boundary points of the function $f\left(x\right)$ are $-2,1,4$ .

Now, locate these boundary points 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 negative into $\left(1,4\right)$ .

Hence, the real values of $x$ is equal to $\left(1,4\right)$ .