Chapter 1.3, Problem 29E

(a)

To determine

To find: The function $f+g$ and its domain.

Answer to Problem 29E

The sum of the functions $\left(f+g\right)\left(x\right)=x^3+5x^2-1$ and its domain is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

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

Calculation:

The sum of the functions $\left(f+g\right)\left(x\right)$ is defined as follows.

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

Substitute $f\left(x\right)$ and $g\left(x\right)$,

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

Notice that $f\left(x\right)+g\left(x\right)$ is a polynomial function.

Since polynomial function is defined on real line, the domain of $\left(f+g\right)\left(x\right)$ is a set of all real numbers.

Thus, $\left(f+g\right)\left(x\right)=x^3+5x^2-1$ whose domain is $ℝ$.

(b)

To determine

To find: The function $f-g$ and its domain.

Answer to Problem 29E

The subtraction of functions $\left(f-g\right)\left(x\right)=x^3-x^2+1$ and its domain is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

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

Calculation:

The function $\left(f-g\right)\left(x\right)$ is defined as follows.

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

Substitute $f\left(x\right)$ and $g\left(x\right)$,

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

Notice that $\left(f-g\right)\left(x\right)$ is a polynomial function.

Since polynomial function is defined on real line, the domain of $\left(f-g\right)\left(x\right)$ is a set of all real numbers.

Thus, $\left(f-g\right)\left(x\right)=x^3-x^2+1$ whose domain is $ℝ$.

(c)

To determine

To find: The function $fg$ and its domain.

Answer to Problem 29E

The product function $\left(fg\right)\left(x\right)=3x^5+6x^4-x^3-2x^2$ and its domain is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

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

Calculation:

The product function $\left(fg\right)\left(x\right)$ is defined as follows.

$\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$

Substitute $f\left(x\right)$ and $g\left(x\right)$,

$\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(x^3+2x^2\right)\left(3x^2-1\right)\\ =\left(3x^2\right)x^3-x^3+2x^2\left(3x^2\right)-2x^2\\ =3x^5-x^3+6x^4-2x^2\\ =3x^5+6x^4-x^3-2x^2\end{array}$

Notice that $\left(fg\right)\left(x\right)$ is a polynomial function.

Since polynomial function is defined on real line, the domain of $\left(fg\right)\left(x\right)$ is a set of all real numbers.

Thus, $\left(fg\right)\left(x\right)=3x^5+6x^4-x^3-2x^2$ whose domain is $ℝ$.

(d)

To determine

To find: The function $\frac{f}{g}$ and its domain.

Answer to Problem 29E

The quotient function $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^3+2x^2}{3x^2-1}$ and its domain is $x\in ℝ-\left\{\pm \frac{1}{\sqrt{3}}\right\}$.

Explanation of Solution

Given:

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

Calculation:

The quotient function $\left(\frac{f}{g}\right)\left(x\right)$ is defined as follows.

$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

Substitute $f\left(x\right)$ and $g\left(x\right)$, $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^3+2x^2}{3x^2-1}$.

Thus, the quotient function is defined when the denominator is not equal to 0.

Hence, the function is $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^3+2x^2}{3x^2-1}$ not defined when, $3x^2-1=0$.

That is, $x=\pm \frac{1}{\sqrt{3}}$.

Therefore, the domain of the function is $x\in ℝ-\left\{\pm \frac{1}{\sqrt{3}}\right\}$.

Thus, $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^3+2x^2}{3x^2-1}$ whose domain is $x\in ℝ-\left\{\pm \frac{1}{\sqrt{3}}\right\}$.