Chapter 0, Problem 44RE

a).

To determine

To find: the formula for $f\circ g$ and $g\circ f$ .

Answer to Problem 44RE

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

Explanation of Solution

Given:

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

Calculation:

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

And,

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

b).

To determine

To find: the domain of $f\circ g$ and $g\circ f$ .

Answer to Problem 44RE

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

Explanation of Solution

Given:

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

Calculation:

The function $f\circ g$ gives a real $y$ value for every real $x$ value less than or equal to $1$ , because the term under fourth root must be positive.

Thus the domain of $f\circ g$ is $\left(-\infty ,1\right]$ .

The function $g\circ f$ gives a real $y$ value for every real $x$ value from $0$ to $1$ .

Thus the domain of $g\circ f$ is $\left[0,1\right]$ .

c).

To determine

To find: the range of $f\circ g$ and $g\circ f$ .

Answer to Problem 44RE

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

Explanation of Solution

Given:

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

Calculation:

The value $f\circ g$ takes on every real number greater than or equal to zero.

Thus, the range of $f\circ g$ is $\left[0,\infty \right)$ .

The value $g\circ f$ takes on every real number from $0$ to $1$ .

Thus, the range of $g\circ f$ is $\left[0,1\right]$ .