Question

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

$f\left(x\right)=\sqrt{x-1};g\left(x\right)=\sqrt{4-x}$

Short answer

The value of $(f+g)\left(x\right)=\sqrt{x-1}+\sqrt{4-x}$and the domain of $f+g$is$\left\{x\right|x\ge 1,x\le 4\}$

The value of $(f-g)\left(x\right)=\sqrt{x-1}-\sqrt{4-x}$and the domain of $f-g$is$\left\{x\right|x\ge 1,x\le 4\}$

The value of role="math" localid="1646140892163" $(f·g)\left(x\right)=\sqrt{x-1}·\sqrt{4-x}$and the domain of $f·g$is $\left\{x\right|x\ge 1,x\le 4\}$.

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x-1}}{\sqrt{4-x}}$and the domain of $\frac{f}{g}$is $\left\{x\right|x\ge 1,x\le 4\}$.

The value of$(f+g)\left(3\right)=\sqrt{2}+1$

The value of$(f-g)\left(4\right)=\sqrt{3}$

The value of$(f·g)\left(2\right)=\sqrt{2}$

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

Step-by-step solution

Step 1. Given Information 

In the given problems we have to solve the given functions f and g, find the following. For parts (a)–(d), we have to also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

The given function is$f\left(x\right)=\sqrt{x-1};g\left(x\right)=\sqrt{4-x}$

Step 2. The function f(x) and g(x) tells us to take the square root of x-1 and 4-x. But only non-negative numbers have real square roots, so the expression under the square root (the radicand) must be non-negative (greater than or equal to zero).

This requires that

$$\begin{gathered} x-1\ge 0\mathrm{and}4-x\ge 0 \\ x-1+1\ge 0+1\mathrm{and}4-x-4\ge 0-4 \\ x\ge 1\mathrm{and}-x\ge -4 \\ x\ge 1\mathrm{and}x\le 4 \end{gathered}$$

So the domain of$f\left(x\right)=\left\{x\right|x\ge 1\}$and$g\left(x\right)=\left\{x\right|x\le 4\}$

Part (a) Step 1. We have to find the value of (f+g)(x)

We know that$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)$

Part (a) Step 2. Putting the value of f(x) and g(x)

$(f+g)\left(x\right)=\sqrt{x-1}+\sqrt{4-x}$

The domain of $f+g$ consists of those numbers x that are in the domains of both fandg. Therefore, the domain of $f+g$is$\left\{x\right|x\ge 1,x\le 4\}$

Part (b) Step 1. We have to find the value of (f-g)(x)We know (f-g)(x)=f(x)-g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$(f-g)\left(x\right)=\sqrt{x-1}-\sqrt{4-x}$

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$is $\left\{x\right|x\ge 1,x\le 4\}$.

Part (c) Step 1. We have to find the value of (f·g)(x)We know that (f·g)(x)=f(x)·g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$(f·g)\left(x\right)=\sqrt{x-1}·\sqrt{4-x}$

The domain of $f·g$ consists of those numbers x that are in the domains of both $f\mathrm{and}g$. Therefore, the domain of $f·g$is $\left\{x\right|x\ge 1,x\le 4\}$.

Part (d) Step 1. We have to find the value of fg(x)We know that fg(x)=f(x)g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x-1}}{\sqrt{4-x}}$

Part (d) Step 2. The domain of fg consists of the numbers x for which g(x)≠0 and that are in the domains of both f and g.  

The domain of $\frac{f}{g}$ consists of those numbers x that are in the domains of both fandg.

The domain of $\frac{f}{g}$is $\left\{x\right|x\ge 1,x\le 4\}$

Part (e) Step 1. We have to find the value of (f+g)(3)From the part (a) we know the value of (f+g)(x)=x-1+4-x

Putting $x=3$in the value of $(f+g)\left(x\right)$

$$\begin{gathered} (f+g)\left(3\right)=\sqrt{3-1}+\sqrt{4-3} \\ (f+g)\left(3\right)=\sqrt{2}+\sqrt{1} \\ (f+g)\left(3\right)=\sqrt{2}+1 \end{gathered}$$

Part (f) Step 1. We have to find the value of (f-g)(4)From the part (a) we know the value of (f-g)(x)=x-1-4-x

Putting $x=4$in the value of $(f-g)\left(x\right)$

$$\begin{gathered} (f-g)\left(4\right)=\sqrt{4-1}-\sqrt{4-4} \\ (f-g)\left(4\right)=\sqrt{3}-\sqrt{0} \\ (f-g)\left(4\right)=\sqrt{3} \end{gathered}$$

Part (g) Step 1. We have to find the value of (f·g)(2)From the part (a) we know the value of (f·g)(x)=x-1·4-x

Putting $x=2$in the value of $(f·g)\left(x\right)$

$$\begin{gathered} (f·g)\left(2\right)=\sqrt{2-1}·\sqrt{4-2} \\ (f·g)\left(2\right)=\sqrt{1}·\sqrt{2} \\ (f·g)\left(2\right)=1·\sqrt{2} \\ (f·g)\left(2\right)=\sqrt{2} \end{gathered}$$

Part (h) Step 1. We have to find the value of fg(1)From the part (a) we know the value of fg(x)=x-14-x

Putting $x=1$in the value of $\left(\frac{f}{g}\right)\left(x\right)$

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{\sqrt{1-1}}{\sqrt{4-1}} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{\sqrt{0}}{\sqrt{3}} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{0}{\sqrt{3}} \\ \left(\frac{f}{g}\right)\left(1\right)=0 \end{gathered}$$