Chapter 2.1, Problem 64AYU

a.

To determine

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

Answer to Problem 64AYU

The value of $\left(f+g\right)\left(x\right)$ is $5x-1$ .

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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 $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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

Substitute $2x+1$ for $f\left(x\right)$ and $3x-2$ for $g\left(x\right)$ in the sum formula.

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =2x+1+3x-2\\ =5x-1\end{array}$

Hence, the value of $\left(f+g\right)\left(x\right)$ is $5x-1$ .

As domain of $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ 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 to Problem 64AYU

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

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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 $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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

Substitute $2x+1$ for $f\left(x\right)$ and $3x-2$ for $g\left(x\right)$ in the difference formula.

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

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

As domain of $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ 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 to Problem 64AYU

The value of $\left(fg\right)\left(x\right)$ is $6x^2-x-2$ .

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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 $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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

Substitute $2x+1$ for $f\left(x\right)$ and $3x-2$ for $g\left(x\right)$ in the product formula.

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(2x+1\right)\left(3x-2\right)\\ =6x^2+3x-4x-2\\ =6x^2-x-2\end{array}$

Hence, the value of $\left(fg\right)\left(x\right)$ is $6x^2-x-2$ .

As domain of $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ 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 to Problem 64AYU

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{\left(2x+1\right)}{\left(3x-2\right)}$ and the domain is all the real numbers $x$ except $x=\frac{2}{3}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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 $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

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

Substitute $2x+1$ for $f\left(x\right)$ and $3x-2$ for $g\left(x\right)$ in the division formula.

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

Find the domain of the function.

  $\begin{array}{c}3x-2=0\\ x=\frac{2}{3}\end{array}$

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

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{\left(2x+1\right)}{\left(3x-2\right)}$ and the domain is all the real numbers $x$ except $x=\frac{2}{3}$ .

e)

To determine

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

Answer to Problem 64AYU

The value of $\left(f+g\right)\left(3\right)$ is $14$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

Calculation:

Now, from part (a)

  $\left(f+g\right)\left(x\right)=5x-1$

Substitute $3$ for $x$ in the composite function $\left(f+g\right)\left(x\right)=5x-1$ .

  $\begin{array}{c}\left(f+g\right)\left(x\right)=5x-1\\ \left(f+g\right)\left(3\right)=5\left(3\right)-1\\ =15-1\\ =14\end{array}$

Hence, the value of $\left(f+g\right)\left(3\right)$ is $14$ .

f)

To determine

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

Answer to Problem 64AYU

The value of $\left(f-g\right)\left(4\right)$ is $-1$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

Calculation:

Now, from part (b)

  $\left(f-g\right)\left(x\right)=-x+3$

Substitute $4$ for $x$ in the composite function $\left(f-g\right)\left(x\right)=-x+3$ .

  $\begin{array}{c}\left(f+g\right)\left(x\right)=-x+3\\ \left(f-g\right)\left(4\right)=-4+3\\ =-1\end{array}$

Hence, the value of $\left(f-g\right)\left(4\right)$ is $-1$ .

g)

To determine

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

Answer to Problem 64AYU

The value of $\left(fg\right)\left(2\right)$ is $20$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

Calculation:

Now, from part (c)

  $\left(fg\right)\left(x\right)=6x^2-x-2$

Substitute $2$ for $x$ in the composite function $\left(fg\right)\left(x\right)=6x^2-x-2$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=6x^2-x-2\\ \left(fg\right)\left(2\right)=6{\left(2\right)}^2-2-2\\ =24-4\\ =20\end{array}$

Hence, the value of $\left(fg\right)\left(2\right)$ is $20$ .

h)

To determine

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

Answer to Problem 64AYU

The value of $\left(\frac{f}{g}\right)\left(1\right)$ is $3$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=2x+1$ and $g\left(x\right)=3x-2$ .

Calculation:

Now, from part (d)

  $\left(\frac{f}{g}\right)\left(x\right)=\frac{\left(2x+1\right)}{\left(3x-2\right)}$

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

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

Hence, the value of $\left(\frac{f}{g}\right)\left(1\right)$ is $3$ .