Chapter 7.1, Problem 62HP

(a)

To determine

State whether the statement is sometimes, always , or never true .

Answer to Problem 62HP

Sometimes

Explanation of Solution

Given:

The domain of two functions f(x) and g(x) that are composed $g\left[f(x)\right]$ is restricted by the domain of f(x).

Calculation:

Let

  $\begin{array}{l}f(x)=\sqrt{x}\\ g(x)=3x\end{array}$

  $g\left[f(x)\right]=g\left(\sqrt{x}\right)=3\sqrt{x}$

Here, the domain of f and $g\left[f(x)\right]$ is all non-negative real numbers, since the radicand of a square root is always non-negative and domain of g is all real numbers.

  $D\left(f\right)=\left\{x\in ℝ:x\ge 0\right\}$ , $D\left(g\right)=\left\{x:x\in ℝ\right\}$ and $D\left(g\circ f\right)=\left\{x\in ℝ:x\ge 0\right\}$

So, in this case , the domain of $g\left[f(x)\right]$ is restricted by the domain of f(x).

Let

  $\begin{array}{l}f(x)=x\\ g(x)=2x\end{array}$

  $g\left[f(x)\right]=g\left(x\right)=2x$

Here, the domain of f and g is all real numbers,and the domain of $g\left[f(x)\right]$ is also all real numbers .

  $D(f)=\left\{x:x\in ℝ\right\}$ , $D(g)=\left\{x:x\in ℝ\right\}$ and $D(g\circ f)=\left\{x:x\in ℝ\right\}$

So, here the domain of $g\left[f(x)\right]$ is same as the domain of f and g .

Let

  $\begin{array}{l}f(x)=\sqrt{x}\\ g(x)=x^2\end{array}$

  $g\left[f(x)\right]=g\left(\sqrt{x}\right)={\left(\sqrt{x}\right)}^2=x$

Here, the domain of f is all non-negative real numbers, since the radicand of a square root is always non-negative and domain of g and $g\left[f(x)\right]$ is all real numbers.

  $D(f)=\left\{x\in ℝ:x\ge 0\right\}$ , $D(g)=\left\{x:x\in ℝ\right\}$ and $D(g\circ f)=\left\{x:x\in ℝ\right\}$

So, here the domain of $g\left[f(x)\right]$ is not restricted by the domain of f .

Hence, the statement is sometimes true.

(b)

To determine

State whether the statement is sometimes, always , or never true .

Answer to Problem 62HP

Sometimes

Explanation of Solution

Given:

The domain of two functions f(x) and g(x) that are composed $g\left[f(x)\right]$ is restricted by the domain of g(x).

Calculation:

Let

  $\begin{array}{l}f(x)=\sqrt{x}\\ g(x)=3x\end{array}$

  $g\left[f(x)\right]=g\left(\sqrt{x}\right)=3\sqrt{x}$

Here, the domain of f and $g\left[f(x)\right]$ is all non-negative real numbers, since the radicand of a square root is always non-negative and domain of g is all real numbers.

  $D\left(f\right)=\left\{x\in ℝ:x\ge 0\right\}$ , $D\left(g\right)=\left\{x:x\in ℝ\right\}$ and $D\left(g\circ f\right)=\left\{x\in ℝ:x\ge 0\right\}$

So, in this case , the domain of $g\left[f(x)\right]$ is restricted by the domain of g(x).

Let

  $\begin{array}{l}f(x)=x\\ g(x)=2x\end{array}$

  $g\left[f(x)\right]=g\left(x\right)=2x$

Here, the domain of f and g is all real numbers,and the domain of $g\left[f(x)\right]$ is also all real numbers .

  $D(f)=\left\{x:x\in ℝ\right\}$ , $D(g)=\left\{x:x\in ℝ\right\}$ and $D(g\circ f)=\left\{x:x\in ℝ\right\}$

So, here the domain of $g\left[f(x)\right]$ is same as the domain of f and g .

Let

  $\begin{array}{l}f(x)=x^2\\ g(x)=\sqrt{x}\end{array}$

  $g\left[f(x)\right]=g\left(x^2\right)=\sqrt{x^2}=\pm x$

Here, the domain of g is all non-negative real numbers, since the radicand of a square root is always non-negative and domain of f and $g\left[f(x)\right]$ is all real numbers.

  $D(f)=\left\{x:x\in ℝ\right\}$ , $D(g)=\left\{x\in ℝ:x\ge 0\right\}$ and $D(g\circ f)=\left\{x:x\in ℝ\right\}$

So, here the domain of $g\left[f(x)\right]$ is not restricted by the domain of gs .

Hence, the statement is sometimes true.