Chapter 4, Problem 1CRT

a.

To determine

Find the domain of $f$ .

Answer to Problem 1CRT

  $\boxed{-\infty ,\infty }$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$ find the domain of $f$ .

Calculation:

Consider the function,

  $f\left(x\right)=x^2-4x$

Determine the values that would make the function undefined. Because there are no values make the function undefined,

So, domain of the function is $\boxed{-\infty ,\infty }$ .

Hence, the solution is, $\boxed{-\infty ,\infty }$

b.

To determine

Find the domain of $g$ .

Answer to Problem 1CRT

  $\boxed{-4,\infty }$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$ find the domain of $g$ .

Calculation:

Consider the function,

  $g\left(x\right)=\sqrt{x+4}$

Determine the values that would make the function undefined. It is a function with square root so, the value under radical is always positive.

  $\begin{array}{l}x+4>0\\ x+4-4>0-4\\ x>-4\end{array}$

We conclude that $x$ cannot less than -4. So domain of the function is $\boxed{-4,\infty }$

Hence, the solution is, $\boxed{-4,\infty }$

c.

To determine

Find $f\left(-2\right),f\left(0\right),f\left(4\right),g\left(0\right),g\left(8\right),g\left(-6\right)$ .

Answer to Problem 1CRT

  $\boxed{12},\boxed{0},\boxed{0},\boxed{2},\boxed{2\sqrt{3}},\boxed{\sqrt{-2}}$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$

  $f\left(-2\right),f\left(0\right),f\left(4\right),g\left(0\right),g\left(8\right),g\left(-6\right)$

Calculation:

Consider the function,

  $f\left(x\right)=x^2-4x$

Put $-2$ in $f\left(x\right)$ ,

  $\begin{array}{l}f\left(-2\right)={\left(-2\right)}^2-4\left(-2\right)\\ =4+8\\ =\boxed{12}\end{array}$

Determine $f\left(0\right)$ put $0$ in $f\left(x\right)$

  $\begin{array}{l}f\left(0\right)={\left(0\right)}^2-4\left(0\right)\\ =0\\ =\boxed{0}\end{array}$

Determine $f\left(4\right)$ ,

  $\begin{array}{l}f\left(4\right)={\left(4\right)}^2-4\left(4\right)\\ =16-16\\ =\boxed{0}\end{array}$

Determine $g\left(0\right)$ ,

  $\begin{array}{l}g\left(0\right)=\sqrt{0+4}\\ =\sqrt{4}\\ =\boxed{2}\end{array}$

Determine $g\left(8\right)$

  $\begin{array}{l}g\left(8\right)=\sqrt{8+4}\\ =\sqrt{12}\\ =\boxed{2\sqrt{3}}\end{array}$

Determine $g\left(-6\right)$

  $\begin{array}{l}g\left(-6\right)=\sqrt{-6+4}\\ =\boxed{\sqrt{-2}}\end{array}$

Hence, the solution is, $\boxed{12},\boxed{0},\boxed{0},\boxed{2},\boxed{2\sqrt{3}},\boxed{\sqrt{-2}}$

d.

To determine

Find $f\left(x+2\right),g\left(x+2\right),f\left(2+h\right)$ .

Answer to Problem 1CRT

  $\boxed{x^2-4},\boxed{\sqrt{x+6}},\boxed{h^2-4}$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$

  $f\left(x+2\right),g\left(x+2\right),f\left(2+h\right)$

Calculation:

Consider the function,

  $f\left(x\right)=x^2-4x$

Determine $f\left(x+2\right)$ put $\left(x+2\right)$ in $f\left(x\right)$

  $\begin{array}{l}f\left(x+2\right)={\left(x+2\right)}^2-4\left(x+2\right)\\ =x^2+4x+4-4x-8\\ =\boxed{x^2-4}\end{array}$

Similarly determine $g\left(x+2\right)$

  $\begin{array}{l}g\left(x+2\right)=\sqrt{x+2+4}\\ =\boxed{\sqrt{x+6}}\end{array}$

Similarly determine $f\left(2+h\right)$ ,

  $\begin{array}{l}f\left(2+h\right)={\left(2+h\right)}^2-4\left(2+h\right)\\ =h^2+4h+4-4h-8\\ =\boxed{h^2-4}\end{array}$

Hence, the solution is, $\boxed{x^2-4},\boxed{\sqrt{x+6}},\boxed{h^2-4}$

e.

To determine

The average rate of change of $g$ between $x=5$ and $x=21$ .

Answer to Problem 1CRT

  $\boxed{22}$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$

The average rate of change of $g$ between $x=5$ and $x=21$ .

Calculation:

Consider the formula for average rate of change.

Average rate of change = $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Now find $f\left(5\right)$ and $f\left(21\right)$

  $f\left(x\right)=x^2-4x$

  $\begin{array}{l}f\left(5\right)=5^2-4\times 5\\ =25-20\\ =5\end{array}$

  $\begin{array}{l}f\left(21\right)={21}^2-4\times 21\\ =357\end{array}$

  $\frac{f\left(21\right)-f\left(5\right)}{21-5}=\frac{357-\left(5\right)}{6}$

  $=\boxed{22}$

Hence, the solution is, $\boxed{22}$

f.

To determine

Find $f\circ g,g\circ f,f\left(g\left(12\right)\right),g\left(f\left(12\right)\right)$ .

Answer to Problem 1CRT

  $\boxed{x+4-4\sqrt{x+4}},\boxed{x-2},\boxed{12},\boxed{10}$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$

  $f\circ g,g\circ f,f\left(g\left(12\right)\right),g\left(f\left(12\right)\right)$ .

Calculation:

Consider $f\circ g,g\circ f,f\left(g\left(12\right)\right),g\left(f\left(12\right)\right)$

  $\begin{array}{l}f\circ g=\left(f\circ g\right)\left(x\right)\left[\therefore f\circ g=f\left(g\right)\right]\\ =f\left(g\left(x\right)\right)\\ =f\left(\sqrt{x+4}\right)\\ f\left(x\right)=x^2-4x\\ put,\\ x=\sqrt{x+4}\\ ={\left(\sqrt{x+4}\right)}^2-4\sqrt{x+4}\\ =\boxed{x+4-4\sqrt{x+4}}\end{array}$

Now,

  $\begin{array}{l}g\circ f=\left(g\circ f\right)\left(x\right)\left[\therefore f\circ g=f\left(g\right)\right]\\ =g\left(f\left(x\right)\right)\\ =g\left(x^2-4x\right)\\ g\left(x\right)=\sqrt{x^2-4x+4}\\ =\sqrt{{\left(x-2\right)}^2}\\ =\boxed{x-2}\end{array}$

  $\begin{array}{l}f\left(g\left(12\right)\right)=12+4-\sqrt{12+4}\\ =16-4\\ =\boxed{12}\\ =g\left(f\left(12\right)\right)=12-2\\ =\boxed{10}\end{array}$

Hence, the solution is, $\boxed{x+4-4\sqrt{x+4}},\boxed{x-2},\boxed{12},\boxed{10}$

g.

To determine

Find the inverse of $g$ .

Answer to Problem 1CRT

  $\boxed{g^{-1}\left(x\right)=x^2-4,x\ge 0}$

Explanation of Solution

Given information:

Let $f\left(x\right)=x^2-4x$ and $g\left(x\right)=\sqrt{x+4}$

Find the inverse of $g$ .

Calculation:

Consider $g\left(x\right)=\sqrt{x+4}$

Assume $g\left(x\right)=y$ so, $g\left(x\right)\ge 0\Rightarrow y\ge 0$

Now, calculate the inverse of $g$ .

  $\begin{array}{l}y=\sqrt{x+4}\\ y^2={\left(\sqrt{x+4}\right)}^2\\ y^2=x+4\\ y^2-4=x\\ hence,\\ g\left(x\right)=y\\ x=g^{-1}y\\ \boxed{g^{-1}\left(x\right)=x^2-4,x\ge 0}\end{array}$

Hence, the solution is, $\boxed{g^{-1}\left(x\right)=x^2-4,x\ge 0}$