Chapter 1.5, Problem 45E

(a)

To determine

To calculate: To determine domain of $f$

Answer to Problem 45E

Domain of f is set of all real numbers such that

  $x\ge 7$

Explanation of Solution

Given:

  $f(x)=\sqrt{x-7}$

Formula Used:

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

Domain of $f$ is calculated as:

Given $f(x)=\sqrt{x-7}$

For $f(x)$ to be defined,

  $x-7\ge 0$

i.e.

  $x>=7$

Therefore, Domain of f is set of all real numbers such that

  $x\ge 7$

Conclusion:

Hence Domain of f is set of all real numbers such that

  $x\ge 7$

(b)

To determine

To calculate: To determine domain of $g$

Answer to Problem 45E

Domain of g is set of all real numbers.

Explanation of Solution

Given:

  $g(x)=4x^2$

Formula Used:

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

Domain of $g$ is calculated as:

Given $g(x)=4x^2$

Now clearly, $g(x)$ is defined for all values of $x$ .

Therefore, domain of g is set of all real numbers.

Conclusion:

Hence domain of g is set of all real numbers.

(c)

To determine

To calculate: To determine domain of $f\circ g$

Answer to Problem 45E

Domain of $f\circ g$ is all real x such that

  $x\le -\frac{\sqrt{7}}{2}\cup x\ge \frac{\sqrt{7}}{2}$

Explanation of Solution

Given:

  $f(x)=\sqrt{x-7}$

  $g(x)=4x^2$

Formula Used:

1. Composition of functions is when one function is inside of another function.

The notation used for the composition of functions looks like this, $\left(f\circ g\right)\left(x\right)$ .

The composition of the function f with g is defined as follows:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ , notice that in the case the function g is inside of the function

2. The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

  $f\circ g$ is calculated as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute: $g(x)=4x^2$

  $\left(f\circ g\right)\left(x\right)$ = $f(4x^2)$

Also, given $f(x)=\sqrt{x-7}$

Thus, $\left(f\circ g\right)\left(x\right)$ is calculated as

  $\left(f\circ g\right)\left(x\right)=\sqrt{4x^2-7}$

  $\left(f\circ g\right)\left(x\right)$ is defined only if:

  $\begin{array}{l}4x^2-7\ge 0\\ 4x^2\ge 7\\ x^2\ge \frac{7}{4}\\ x\le -\frac{\sqrt{7}}{2}\cup x\ge \frac{\sqrt{7}}{2}\end{array}$

Domain of $f\circ g$ is set of all real $x$ such that

  $x\le -\frac{\sqrt{7}}{2}\cup x\ge \frac{\sqrt{7}}{2}$

Conclusion:

Hence Domain of $f\circ g$ is set of all real $x$ such that

  $x\le -\frac{\sqrt{7}}{2}\cup x\ge \frac{\sqrt{7}}{2}$