Chapter 1.3, Problem 30E

(a)

To determine

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

Answer to Problem 30E

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

Explanation of Solution

Given:

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

Result used:

The sum of the functions $f\left(x\right)\text{ and g}\left(x\right)$ is defined as follows.

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

Calculation:

Substitute the value of $f\left(x\right)$ and $g\left(x\right)$ in $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$.

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

As the radicand cannot take negative values, $3-x>0$.

On simplification,

$\begin{array}{c}3-x>0\\ x<3\end{array}$

Also $x^2-1>0$.

Simplify the above inequality as follows.

$\begin{array}{c}{x}^2-1>0\\ x^2>1\\ x>\pm 1\end{array}$

So the value of x lies in the interval $x<3,x>\pm 1$.

Hence, the domain of $\sqrt{3-x}$ is $\left(-\infty ,3\right]$ and the domain of $\sqrt{x^2-1}$ is $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$.

So the combined domain is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

Therefore, the domain of $f\left(x\right)+g\left(x\right)=\sqrt{3-x}+\sqrt{x^2-1}$ is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

(b)

To determine

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

Answer to Problem 30E

The function $f-g$ is $f\left(x\right)-g\left(x\right)=\sqrt{3-x}-\sqrt{x^2-1}$ and its domain is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

Explanation of Solution

Result used:

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

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

Calculation:

Substitute the value of $f\left(x\right)$ and $g\left(x\right)$ in $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$.

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

From part (a), the domain of the function $f+g$ is $\left(-\infty ,-1\right]\cup \left[1,3\right]$, which is also the domain of $f-g$.

Therefore, the value of the given function is $f\left(x\right)-g\left(x\right)=\sqrt{3-x}-\sqrt{x^2-1}$ whose domain is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

(c)

To determine

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

Answer to Problem 30E

The product function $f\left(x\right)\cdot g\left(x\right)=\left(\sqrt{3-x}\right)\cdot \left(\sqrt{x^2-1}\right)$ and its domain is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

Explanation of Solution

Result used:

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

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

Calculation:

Substitute the value of $f\left(x\right)$ and $g\left(x\right)$ in $\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$.

$f\left(x\right)\cdot g\left(x\right)=\left(\sqrt{3-x}\right)\cdot \left(\sqrt{x^2-1}\right)$.

From part (a), the domain of the function $f+g$ is $\left(-\infty ,-1\right]\cup \left[1,3\right]$, which is also the domain of $fg$.

Therefore, the value of the given function is $f\left(x\right)\cdot g\left(x\right)=\left(\sqrt{3-x}\right)\cdot \left(\sqrt{x^2-1}\right)$ whose domain is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

(d)

To determine

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

Answer to Problem 30E

The quotient function is $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(\sqrt{3-x}\right)}{\left(\sqrt{x^2-1}\right)}$ and its domain is $\left(-\infty ,-1\right)\cup \left(1,3\right]$.

Explanation of Solution

Result used:

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

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

Calculation:

Substitute the value of $f\left(x\right)$ and $g\left(x\right)$ in $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$, $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(\sqrt{3-x}\right)}{\left(\sqrt{x^2-1}\right)}$.

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

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

Simplify the equation $x^2-1=0$.

$\begin{array}{c}{x}^2-1=0\\ x^2=1\\ x=\pm 1\end{array}$

So, the numbers −1 and 1 are excluded from the domain.

From part (a), the domain of the function $f+g$ is $\left(-\infty ,-1\right]\cup \left[1,3\right]$.

Remove −1 and 1 from $\left(-\infty ,-1\right]\cup \left[1,3\right]$, $\left(-\infty ,-1\right)\cup \left(1,3\right]$.

Therefore, the domain of the function is $\left(-\infty ,-1\right)\cup \left(1,3\right]$.

Therefore, the value of the given function is $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(\sqrt{3-x}\right)}{\left(\sqrt{x^2-1}\right)}$ whose domain is $\left(-\infty ,-1\right)\cup \left(1,3\right]$.