Q. 70 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

Step-by-Step Solution

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 $(f·g)\left(x\right)=\sqrt{x-1}·\sqrt{4-x}$ 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 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 f and g. 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 f g ( x ) We know that f g ( x ) = f ( x ) g ( x )

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

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

Part (d) Step 2. The domain of f g 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 f and g. 

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 f g ( 1 ) From the part (a) we know the value of f g ( x ) = x - 1 4 - x

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

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