Chapter 2.1, Problem

a.

To determine

The value of $\left(f+g\right)\left(x\right)$ and the domain.

Answer

The value of $\left(f+g\right)\left(x\right)$ is $2x^2+x-1$ .

Domain of $\left(f+g\right)\left(x\right)$ is set of all real numbers.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Formula Used:

Let f and g be two functions.

Use the sum formula, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

For a composite function $f+g$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Use the sum formula, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

Substitute $x-1$ for $f\left(x\right)$ and $2x^2$ for $g\left(x\right)$ in the sum formula.

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ x-1+2x^2\\ =2x^2+x-1\end{array}$

Hence, the value of $\left(f+g\right)\left(x\right)$ is $2x^2+x-1$ .

As domain of $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ is set of real numbers.

So, domain of $f+g$ is set of all real numbers.

b.

To determine

The value of $\left(f-g\right)\left(x\right)$ and its domain.

Answer

The value of $\left(f-g\right)\left(x\right)$ is $x-2x^2-1$ .

Domain of $\left(f-g\right)\left(x\right)$ is set of all real numbers.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Formula Used:

Let f and g be two functions.

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

For a composite function $f-g$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

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

Substitute $x-1$ for $f\left(x\right)$ and $2x^2$ for $g\left(x\right)$ in the difference formula.

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =x-1-2x^2\\ =x-2x^2-1\end{array}$

Hence, the value of $\left(f-g\right)\left(x\right)$ is $x-2x^2-1$ .

As domain of $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ is set of real numbers.

So, domain of $f-g$ is set of all real numbers.

c.

To determine

The value of $\left(fg\right)\left(x\right)$ and its domain.

Answer

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

Domain of $\left(fg\right)\left(x\right)$ is set of all real numbers.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Formula Used:

Let f and g be two functions.

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

For a composite function $fg$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

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

Substitute $x-1$ for $f\left(x\right)$ and $2x^2$ for $g\left(x\right)$ in the product formula.

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(x-1\right)2x^2\\ =2x^3-2x^2\end{array}$

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

As domain of $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ is set of real numbers.

So, domain of $\left(fg\right)\left(x\right)$ is set of all real numbers.

d.

To determine

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

Answer

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{x-1}{2x^2}$ and the domain is all the real numbers $x$ except $x=0$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Formula Used:

Let f and g be two functions.

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

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

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

Substitute $x-1$ for $f\left(x\right)$ and $2x^2$ for $g\left(x\right)$ in the division formula.

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{x-1}{2x^2}\end{array}$

Find the domain of the function.

  $\begin{array}{c}2x^2=0\\ x=0\end{array}$

The domain is all the real numbers $x$ except $x=\frac{2}{3}$ because the denominator becomes zero at $x=0$ .

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{x-1}{2x^2}$ and the domain is all the real numbers $x$ except $x=0$ .

e)

To determine

The value of $\left(f+g\right)\left(3\right)$ .

Answer

The value of $\left(f+g\right)\left(3\right)$ is $20$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Calculation:

Now, from part (a)

  $\left(f+g\right)\left(x\right)=2x^2+x-1$

Substitute $3$ for $x$ in the composite function $\left(f+g\right)\left(x\right)=2x^2+x-1$ .

  $\begin{array}{c}\left(f+g\right)\left(x\right)=2x^2+x-1\\ \left(f+g\right)\left(3\right)=2{\left(3\right)}^2+3-1\\ =18+3-1\\ =20\end{array}$

Hence, the value of $\left(f+g\right)\left(3\right)$ is $20$ .

f)

To determine

The value of $\left(f-g\right)\left(4\right)$ .

Answer

The value of $\left(f-g\right)\left(4\right)$ is $-29$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Calculation:

Now, from part (b)

  $\left(f-g\right)\left(x\right)=x-2x^2-1$

Substitute $4$ for $x$ in the composite function $\left(f-g\right)\left(x\right)=x-2x^2-1$ .

  $\begin{array}{c}\left(f-g\right)\left(x\right)=x-2x^2-1\\ \left(f-g\right)\left(4\right)=4-2{\left(4\right)}^2-1\\ =4-32-1\\ =-29\end{array}$

Hence, the value of $\left(f-g\right)\left(4\right)$ is $-29$ .

g)

To determine

The value of $\left(fg\right)\left(2\right)$ .

Answer

The value of $\left(fg\right)\left(2\right)$ is $8$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Calculation:

Now, from part (c)

  $\left(fg\right)\left(x\right)=2x^3-2x^2$

Substitute $2$ for $x$ in the composite function $\left(fg\right)\left(x\right)=2x^3-2x^2$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=2x^3-2x^2\\ \left(fg\right)\left(2\right)=2{\left(2\right)}^3-2{\left(2\right)}^2\\ =16-8\\ =8\end{array}$

Hence, the value of $\left(fg\right)\left(2\right)$ is $8$ .

h)

To determine

The value of $\left(\frac{f}{g}\right)\left(1\right)$ .

Answer

The value of $\left(\frac{f}{g}\right)\left(1\right)$ is $0$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=x-1$ and $g\left(x\right)=2x^2$ .

Calculation:

Now, from part (d)

  $\left(\frac{f}{g}\right)\left(x\right)=\frac{x-1}{2x^2}$

Substitute $1$ for $x$ in the composite function $\left(\frac{f}{g}\right)\left(x\right)=\frac{x-1}{2x^2}$ .

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{x-1}{2x^2}\\ \left(\frac{f}{g}\right)\left(1\right)=\frac{\left(1-1\right)}{2{\left(1\right)}^2}\\ =0\end{array}$

Hence, the value of $\left(\frac{f}{g}\right)\left(1\right)$ is $0$ .