Chapter 1, Problem 44RE

(a)

To determine

To find: The formula for the functions $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 44RE

The value of the function $\left(f\circ g\right)\left(x\right)$ is $\sqrt[4]{1-x}$ and the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{1-\sqrt{x}}$ .

Explanation of Solution

Given information: The given functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{1-x}$ .

Calculation:

The formula for the function $\left(f\circ g\right)\left(x\right)$ can be calculated as:

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\sqrt{1-x}\right)\end{array}$

Substitute $\sqrt{1-x}$ for $x$ in the function $f\left(x\right)=\sqrt{x}$ .

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=\sqrt{\sqrt{1-x}}\\ ={\left[{\left(1-x\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}\\ ={\left(1-x\right)}^{\frac{1}{4}}\\ =\sqrt[4]{1-x}\end{array}$

So, the formula for the function $\left(f\circ g\right)\left(x\right)$ is $\sqrt[4]{1-x}$ .

The formula for the function $\left(g\circ f\right)\left(x\right)$ can be calculated as:

  $\begin{array}{c}\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\sqrt{x}\right)\end{array}$

Substitute $\sqrt{x}$ for $x$ in the function $g\left(x\right)=\sqrt{1-x}$ .

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

Therefore, the value of the function $\left(f\circ g\right)\left(x\right)$ is $\sqrt[4]{1-x}$ and the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{1-\sqrt{x}}$ .

(b)

To determine

To find: The domain of the function of $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 44RE

The domain of the function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,1\right]$ and the function $\left(g\circ f\right)\left(x\right)$ is $\left[0,1\right]$ .

Explanation of Solution

Given information: The given functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{1-x}$ .

Calculation:

From part (a), the formula of the function $\left(f\circ g\right)\left(x\right)$ is $\sqrt[4]{1-x}$ and the formula of the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{1-\sqrt{x}}$ .

The domain of the function $\left(f\circ g\right)\left(x\right)$ is:

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

And

  $\begin{array}{c}1-x\ge 0\\ x\le 1\end{array}$

So, the domain of the function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,1\right]$ .

The domain of the function $\left(g\circ f\right)\left(x\right)=\sqrt{1-\sqrt{x}}$ is $\left[0,1\right]$ .

Therefore, the domain of the function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,1\right]$ and domain of the function $\left(g\circ f\right)\left(x\right)$ is $\left[0,1\right]$ .

(c)

To determine

To find: The range of the function of $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 44RE

The range of the function $\left(f\circ g\right)\left(x\right)$ is $\left[0,\infty \right)$ and the range of function $\left(g\circ f\right)\left(x\right)$ is $\left[0,1\right]$ .

Explanation of Solution

Given information: The given functions are $f(x)=\sqrt{x}$ and $g(x)=\sqrt{1-x}$ .

Calculation:

From part (a), the formula of the function $\left(f\circ g\right)\left(x\right)$ is $\sqrt[4]{1-x}$ and the formula of the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{1-\sqrt{x}}$ .

The range of the function $\left(f\circ g\right)\left(x\right)=\sqrt[4]{1-x}$ is $\left[0,\infty \right)$ . The range of the function $\left(g\circ f\right)\left(x\right)=\sqrt{1-\sqrt{x}}$ is $\left[0,1\right]$ .

Therefore, the range of the function $\left(f\circ g\right)\left(x\right)$ is $\left[0,\infty \right)$ and the range of function $\left(g\circ f\right)\left(x\right)$ is $\left[0,1\right]$ .