Chapter 5, Problem 1CT

(a)

To determine

The composite function $f\circ g$ and its domain, where $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

Answer to Problem 1CT

Solution:

The composite function $f\circ g$ is $\frac{2x+7}{2x+3}$ .

The domain of $f\circ g$ is the set $\left(-\infty ,-\frac{3}{2}\right)\cup \left(-\frac{3}{2},\infty \right)$ .

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

The definition of a composite function is $\left(f\circ g\right)\left(x\right)=f\left(g(x)\right)$ .

Plug in $g\left(x\right)=2x+5$ ,

  $\Rightarrow \left(f\circ g\right)\left(x\right)=f\left(2x+5\right)$

To find $f\left(2x+5\right)$ , replace $x$ in $f\left(x\right)=\frac{x+2}{x-2}$ by $2x+5$ ,

  $\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{\left(2x+5\right)+2}{\left(2x+5\right)-2}$

  $\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{2x+5+2}{2x+5-2}$

  $\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{2x+7}{2x+3}$

Hence, the composite function is $\left(f\circ g\right)\left(x\right)=\frac{2x+7}{2x+3}$ .

Now, to find the domain of $\left(f\circ g\right)\left(x\right)$ ,

The domain of the composite function $f\circ g$ is the set of all numbers $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$ .

First, find the domain of $g$ .

The function $g\left(x\right)=2x+5$ is a polynomial function.

Polynomial is defined everywhere.

Thus, the domain of the function $g\left(x\right)=2x+5$ is $\left(-\infty ,\infty \right)$ .

Next, find the domain of $f$ .

The function $f\left(x\right)=\frac{x+2}{x-2}$ is a rational function.

The rational function is undefined at the point where the denominator is zero.

Here, the denominator of $f\left(x\right)$ is $x-2$ .

Equate denominator $x-2$ equal to zero,

  $\Rightarrow x-2=0$

  $\Rightarrow x=2$

Therefore, the function $f\left(x\right)=\frac{x+2}{x-2}$ is defined for all real numbers except $x=2$ .

That is, in interval notation $\left(-\infty ,2\right)\cup \left(2,\infty \right)$ .

Thus, the domain of function $f\left(x\right)=\frac{x+2}{x-2}$ is $\left\{x|x\ne 2\right\}$ .

Thus, the domain of $f\left(x\right)=\frac{x+2}{x-2}$ is $\left\{x|x\ne 2\right\}$ , so $2$ is excluded from the domain of $f\circ g$ .

This means that $g\left(x\right)$ cannot be equal to $2$ .

Now, solve the equation $g\left(x\right)=2$ to determine what additional values of $x$ to exclude.

  $2x+5=2$

Subtract $5$ from both sides.

  $2x+5-5=2-5$

  $\Rightarrow 2x=-3$

Divide both sides by $2$ ,

  $\frac{2x}{2}=\frac{-3}{2}$

  $\Rightarrow x=-\frac{3}{2}$

Therefore, exclude $-\frac{3}{2}$ from the domain of $f\circ g$ .

Therefore, the domain of the function $f\circ g$ is $\left\{x|x=-\frac{3}{2}\right\}$ .

Therefore, the domain of the function $f\circ g$ in interval notation is $\left(-\infty ,-\frac{3}{2}\right)\cup \left(-\frac{3}{2},\infty \right)$ .

(b)

To determine

The value of $\left(g\circ f\right)\left(-2\right)$ , where $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

Answer to Problem 1CT

Solution:

The value of $\left(g\circ f\right)\left(-2\right)$ is $5$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

The definition of a composite function is $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$ .

  $\Rightarrow \left(g\circ f\right)\left(-2\right)=g\left(f\left(-2\right)\right)$

As $f\left(x\right)=\frac{x+2}{x-2}$ , then $f\left(-2\right)=\frac{-2+2}{-2-2}=\frac{0}{-4}=0$ .

By substituting the value of $f\left(-2\right)=0$ in the equation $\left(g\circ f\right)\left(-2\right)=g\left(f\left(-2\right)\right)$ ,

  $\Rightarrow \left(g\circ f\right)\left(-2\right)=g\left(0\right)$

As $g\left(x\right)=2x+5$ ,

  $\Rightarrow \left(g\circ f\right)\left(-2\right)=g\left(0\right)$

  $=2\left(0\right)+5$

  $=5$

Thus, the composite function $\left(g\circ f\right)\left(-2\right)$ is $5$ .

(c)

To determine

The value of $\left(f\circ g\right)\left(-2\right)$ , where, $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

Answer to Problem 1CT

Solution:

The composite function $\left(f\circ g\right)\left(-2\right)$ is $-3$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\frac{x+2}{x-2}$ and $g\left(x\right)=2x+5$ .

The definition of the composite function is $\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)$ .

  $\Rightarrow \left(f\circ g\right)\left(-2\right)=f\left(g\left(-2\right)\right)$

As $g\left(x\right)=2x+5$ , then $g\left(-2\right)=2\left(-2\right)+5=-4+5=1$.

By substituting the value of $g\left(-2\right)=1$ in the equation $\left(f\circ g\right)\left(-2\right)=f\left(g\left(-2\right)\right)$ ,

  $\Rightarrow \left(f\circ g\right)\left(-2\right)=f\left(1\right)$

As $f\left(x\right)=\frac{x+2}{x-2}$ ,

  $\Rightarrow \left(f\circ g\right)\left(-2\right)=\frac{1+2}{1-2}$

  $=\frac{3}{-1}$

  $=-3$

  $\Rightarrow \left(f\circ g\right)\left(-2\right)=-3$

Thus, the composite function $\left(f\circ g\right)\left(-2\right)$ is $-3$ .