Chapter 1.5, Problem 46E

(a)

To determine

To calculate: To determine domain of $f$

Answer to Problem 46E

Domain of f is set of all real numbers such that

  $x>=-3$

Explanation of Solution

Given:

  $f(x)=\sqrt{x+3}$

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 $f(x)=\sqrt{x+3}$

For $f(x)$ to be defined,

  $x+3\ge 0$

i.e.

  $x\ge -3$

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

  $x\ge -3$

Conclusion:

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

  $x\ge -3$

(b)

To determine

To calculate: To determine domain of $g$

Answer to Problem 46E

Domain of g is set of all real numbers.

Explanation of Solution

Given:

  $g(x)=\frac{x}{2}$

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 $g(x)=\frac{x}{2}$

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 46E

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

  $x\ge -6$

Explanation of Solution

Given:

  $f(x)=\sqrt{x+3}$

  $g(x)=\frac{x}{2}$

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: $g(x)=\frac{x}{2}$

  $\left(f\circ g\right)\left(x\right)$ = $f(\frac{x}{2})$

Also, given $f(x)=\sqrt{x+3}$

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

  $\left(f\circ g\right)\left(x\right)=\sqrt{\frac{x}{2}+3}$

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

  $\begin{array}{l}\frac{x}{2}+3\ge 0\\ \frac{x}{2}\ge -3\\ x\ge -6\end{array}$

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

  $x\ge -6$

Conclusion:

Hence, domain of $f\circ g$ is set of all real numbers such that

  $x\ge -6$