Chapter 12, Problem 6CR

(a)

To determine

The value of function $\left(f\circ g\right)\left(2\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

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

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

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

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

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

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

As $f\left(x\right)=\frac{3x}{x-2}$, then $f\left(5\right)=\frac{3\left(5\right)}{5-2}=\frac{15}{3}=5$.

Hence, the value of $\left(f\circ g\right)\left(2\right)$ is $5$.

(b)

To determine

The value of the function $\left(g\circ f\right)\left(4\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The value of $\left(g\circ f\right)\left(4\right)$ is $13$.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

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

$\Rightarrow \left(g\circ f\right)\left(4\right)=g\left(f\left(4\right)\right)$

As $f\left(x\right)=\frac{3x}{x-2}$, then $f\left(4\right)=\frac{3\left(4\right)}{4-2}=\frac{12}{2}=6$,

$\Rightarrow \left(g\circ f\right)\left(4\right)=g\left(6\right)$

As $g\left(x\right)=2x+1$, then $g\left(6\right)=2\left(6\right)+1=13$.

Hence, the value of $\left(g\circ f\right)\left(4\right)$ is $13$.

(c)

To determine

The value of the function $\left(f\circ g\right)\left(x\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The value of $\left(f\circ g\right)\left(x\right)$ is $\frac{6x+3}{2x-1}$.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

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

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

As $f\left(x\right)=\frac{3x}{x-2}$, then $f\left(2x+1\right)=\frac{3\left(2x+1\right)}{\left(2x+1\right)-2}=\frac{6x+3}{2x-1}$,

$\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{6x+3}{2x+1}$

Hence, the value of $\left(f\circ g\right)\left(x\right)$ is $\frac{6x+3}{2x-1}$.

(d)

To determine

The domain of $\left(f\circ g\right)\left(x\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The domain of the composite function $\left(f\circ g\right)\left(x\right)$ is $\left\{x|x\ne \frac{1}{2}\right\}$.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

The domain of a function is the set of $x$ values at which the function is real and defined.

As $f\left(x\right)=\frac{3x}{x-2}$ is a rational function, it is not defined at which the denominator becomes zero.

The denominator of $f\left(x\right)$ is $x-2$.

Solve $x-2=0\Rightarrow x=2$.

So, the domain of $f\left(x\right)$ is all values of $x$ except $x=2$. That is, $x\ne 2$.

That is, domain of $f\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$.

Now, $g\left(x\right)=2x+1$ is a polynomial function and defined for all values of $x$.

Therefore, the domain of $g\left(x\right)$ is $\left(-\infty ,\infty \right)$.

The domain of $f\circ g$ is the set of values of $x$ in the domain of $g$ for which $g\left(x\right)$ is in the domain of $f$. That is, exclude those values of $x$ from the domain of $g$ for which $g\left(x\right)$ is not in the domain of $f$.

The domain of $f\left(x\right)$ is $x\ne 2$.

Now, to find domain of $f\left(g\left(x\right)\right)$, solve $g\left(x\right)=2$, and exclude these values from domain.

$\Rightarrow 2x+1=2$

$\Rightarrow 2x=1$

$\Rightarrow x=\frac{1}{2}$

Therefore, the domain of the composite function $\left(f\circ g\right)\left(x\right)$ is $\left\{x|x\ne \frac{1}{2}\right\}$.

(e)

To determine

The value of the function $\left(g\circ f\right)\left(x\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The value of $\left(g\circ f\right)\left(x\right)$ is $\frac{7x-2}{x-2}$.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

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

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

As $g\left(x\right)=2x+1$, then $g\left(\frac{3x}{x-2}\right)=2\left(\frac{3x}{x-2}\right)+1=\frac{6x}{x-2}+1=\frac{6x+x-2}{x-2}=\frac{7x-2}{x-2}$,

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

Hence, the value of $\left(g\circ f\right)\left(x\right)$ is $\frac{7x-2}{x-2}$.

(f)

To determine

The domain of $\left(g\circ f\right)\left(x\right)$, where $f\left(x\right)=\frac{3x}{x-2}$, and $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The domain of the composite function $\left(g\circ f\right)\left(x\right)$ is $\left\{x|x\ne 2\right\}$.

Explanation of Solution

Given information:

The functions $f\left(x\right)=\frac{3x}{x-2}$ and $g\left(x\right)=2x+1$.

Explanation:

The domain of a function is the set of $x$ values at which the function is real and defined.

As $f\left(x\right)=\frac{3x}{x-2}$ is a rational function, it is not defined at which the denominator becomes zero.

The denominator of $f\left(x\right)$ is $x-2$.

Solve $x-2=0\Rightarrow x=2$.

So, the domain of $f\left(x\right)$ is all values of $x$ except $x=2$. That is, $x\ne 2$.

That is, domain of $f\left(x\right)$ is $\left(-\infty ,2\right)\cup \left(2,\infty \right)$.

Now, $g\left(x\right)=2x+1$ is a polynomial function and defined for all values of $x$.

Therefore, the domain of $g\left(x\right)$ is $\left(-\infty ,\infty \right)$.

The domain of $g\circ f$ is the set of values of $x$ in the domain of $f$ for which $f\left(x\right)$ is in the domain of $g$. That is, exclude those values of $x$ from the domain of $f$ for which $f\left(x\right)$ is not in the domain of $g$.

Hence, the domain of $\left(g\circ f\right)\left(x\right)$ is $\left\{x|x\ne 2\right\}$.

(g)

To determine

The inverse of the function $g$ and its domain, where $g\left(x\right)=2x+1$.

Answer to Problem 6CR

Solution:

The inverse of the function $g$ is $g^{-1}\left(x\right)=\frac{x-1}{2}$.

The domain of the inverse function $g^{-1}\left(x\right)$ is the set of all real numbers.

Explanation of Solution

Given information:

The function $g\left(x\right)=2x+1$.

Explanation:

Let $y=g\left(x\right)$,

$\Rightarrow y=2x+1$.

By interchanging the roles of $x$ and $y$, the function becomes $x=2y+1$.

Now, by solving the function for $y$,

$\Rightarrow x-1=2y$

$\Rightarrow \frac{x-1}{2}=y$

$\Rightarrow g^{-1}\left(x\right)=\frac{x-1}{2}$

The domain of a function is the set of $x$ values at which the function is real and defined.

As $g^{-1}\left(x\right)$ is defined for all $x$, then its domain is the set of all real numbers.

(h)

To determine

The inverse of the function $f$ and its domain, where $f\left(x\right)=\frac{3x}{x-2}$.

Answer to Problem 6CR

Solution:

The inverse of the function $f$ is $f^{-1}\left(x\right)=\frac{2x}{x-3}$.

The domain of the inverse function $f^{-1}\left(x\right)$ is $\left\{x|x\ne 3\right\}$.

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{3x}{x-2}$.

Explanation:

Let $y=f\left(x\right)$,

$\Rightarrow y=\frac{3x}{x-2}$

By interchanging the roles of $x$ and $y$, the function becomes $x=\frac{3y}{y-2}$,

Now, by solving the function for $y$,

$\Rightarrow x\left(y-2\right)=3y$

$\Rightarrow xy-2x=3y$

$\Rightarrow xy-3y=2x$

$\Rightarrow y\left(x-3\right)=2x$

$\Rightarrow y=\frac{2x}{x-3}$

Therefore, $f^{-1}\left(x\right)=\frac{2x}{x-3}$.

The domain of a function is the set of $x$ values at which the function is real and defined.

As $f^{-1}\left(x\right)=\frac{2x}{x-3}$ is not defined when its denominator is zero, Solve $x-3=0\Rightarrow x=3$, and exclude this value from the domain of inverse function.

Hence, the domain of $f^{-1}\left(x\right)$ is $\left\{x|x\ne 3\right\}$.