Chapter 2.5, Problem 15E

a.

To determine

To find the possible rational zeros of $f$ , use the Rational Zero Test. Also, check the zeros of $f$ in the graph.

Answer to Problem 15E

The rational solutions for the function $f$ is $-2,-1,1$ .

Explanation of Solution

Given information:

The functionis $f\left(x\right)=x^3+2x^2-x-2$ .

The graph is:

  

Formula Used:

Rational Zero Test.

The formula is: $\text{Possible}\text{rational}\text{zeros}=\frac{\text{Factors}\text{of}\text{constant}\text{term}}{\text{Factors}\text{of}\text{leading}\text{coefficient}}$ .

Calculation:

The function is $f\left(x\right)=x^3+2x^2-x-2$ .

To find the zeros of function, use the Rational Zero Test.

The leading coefficient is 1 and the constant term is $-2$ .

  $\begin{array}{c}\text{Possible}\text{rational}\text{zeros}=\frac{\text{Factors}\text{of}\text{constant}\text{term}}{\text{Factors}\text{of}\text{leading}\text{coefficient}}\\ =\frac{\text{Factors}\text{of}-2}{\text{Factors}\text{of}1}\\ =\frac{\pm 1,\pm 2}{\pm 1}\\ =\pm 1,\pm 2\end{array}$

The polynomial has only three solutions because the highest degree of the polynomial is three.

The rational solutions for the function $f$ is $-2,-1,1$ .

  

Hence, the rational solutions for the function $f$ is $-2,-1,1$ .

b.

To determine

Find the value of $\left(f-g\right)\left(x\right)$ .

Answer to Problem 15E

The value of $\left(f-g\right)\left(x\right)$ is $x^2-4x+5$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Formula Used:

Let f and g be two functions.

Use the difference formula, $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

Calculation:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Use the difference formula, $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

Substitute $x^2$ for $f\left(x\right)$ and $4x-5$ for $g\left(x\right)$ in the difference formula.

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =x^2-\left(4x-5\right)\\ =x^2-4x+5\end{array}$

Hence, the value of $\left(f-g\right)\left(x\right)$ is $x^2-4x+5$ .

c.

To determine

Find the value of $\left(fg\right)\left(x\right)$ .

Answer to Problem 15E

The value of $\left(fg\right)\left(x\right)$ is $4x^3-5x^2$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Formula Used:

Let f and g be two functions.

Use the product formula, $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$ .

Calculation:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Use the product formula, $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$ .

Substitute $x^2$ for $f\left(x\right)$ and $4x-5$ for $g\left(x\right)$ in the product formula.

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(x^2\right)\left(4x-5\right)\\ =4x^3-5x^2\end{array}$

Hence, the value of $\left(fg\right)\left(x\right)$ is $4x^3-5x^2$ .

d.

To determine

Find the value of $\left(\frac{f}{g}\right)\left(x\right)$ . Also, find the domain of the function $\left(\frac{f}{g}\right)$ .

Answer to Problem 15E

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{x^2}{\left(4x-5\right)}$ and the domain is all the real numbers $x$ except $x=\frac{5}{4}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Formula Used:

Let f and g be two functions.

Use the division formula, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

Calculation:

The functions are $f\left(x\right)=x^2$ and $g\left(x\right)=4x-5$ .

Use the division formula, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

Substitute $x^2$ for $f\left(x\right)$ and $4x-5$ for $g\left(x\right)$ in the division formula.

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{x^2}{\left(4x-5\right)}\end{array}$

Find the domain of the function.

  $\begin{array}{c}4x-5=0\\ x=\frac{5}{4}\end{array}$

The domain is all the real numbers $x$ except $x=\frac{5}{4}$ because the denominator becomes zero at $x=2$ .

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{x^2}{\left(4x-5\right)}$ and the domain is all the real numbers $x$ except $x=\frac{5}{4}$ .