Chapter 2.8, Problem 31E

Solution

(a).

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

The real value of $x$ are equal to $-\frac{5}{2},3$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{\left(2x+5\right)\sqrt{x-3}}{{\left(x-4\right)}^2}$

Calculation:

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

  $\begin{array}{c}\left(2x+5\right)\sqrt{x-3}=0\\ 2x+5=0\\ 2x=-5\\ x=-\frac{5}{2}\end{array}$

And

  $\begin{array}{c}\left(2x+5\right)\sqrt{x-3}=0\\ \sqrt{x-3}=0\\ x-3=0\\ x=3\end{array}$

Thus, the real values of $x$ are equal to $-\frac{5}{2},3$ .

(b).

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

The real values of $x$ is equal to $4$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{\left(2x+5\right)\sqrt{x-3}}{{\left(x-4\right)}^2}$

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:

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

Therefore, the real values of $x$ is $4$ 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 $\left(3,4\right)\cup \left(4,\infty \right)$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{\left(2x+5\right)\sqrt{x-3}}{{\left(x-4\right)}^2}$

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:

  $\begin{array}{c}\left(2x+5\right)\sqrt{x-3}=0\\ x=-\frac{5}{2},3\end{array}$

And

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

So, the boundary points of the function $f\left(x\right)$ are $-\frac{5}{2},3,4$ .

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 $\left(3,4\right)\cup \left(4,\infty \right)$ .

Hence, the real values of $x$ is equal to $\left(3,4\right)\cup \left(4,\infty \right)$ .

(d).

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

No real values of $x$ .

Given information:

Consider the given function.

  $f\left(x\right)=\frac{\left(2x+5\right)\sqrt{x-3}}{{\left(x-4\right)}^2}$

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:

  $\begin{array}{c}\left(2x+5\right)\sqrt{x-3}=0\\ x=-\frac{5}{2},3\end{array}$

And

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

So, the boundary points of the function $f\left(x\right)$ are $-\frac{5}{2},3,4$ .

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

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