Q 35. - Precalculus Enhanced with Graphing Utilities

Question

For the given functions f and g, find:

$a) f\circ g b) g\circ f c) f \circ f d) g\circ g$

State the domain of each composite function. 

$f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(a\right) f\circ g=\frac{3x}{2-x} and domain is (-\infty ,0) \cup (0, 2) \cup (2, \infty ) \\ \mathrm{Part} \left(\mathrm{b}\right) g\circ f=\frac{2x-2}{3} and domain is \left(-\infty ,\infty \right) \\ \mathrm{Part} \left(\mathrm{c}\right) f\circ f=\frac{3x-3}{-x+4} and domain is \left(-\infty ,\infty \right) \\ \mathrm{Part} \left(\mathrm{d}\right) g\circ g=x and domain is \left(-\infty ,\infty \right) \end{gathered}$$

Part (a) Step 1. Given information.

We have given,

$f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$

Part (a) Step 2. Concept used.

A function which is depends on any other function we can call it as composite function.

$f\hspace{0.17em}\circ \hspace{0.17em}g=f\left(g\left(x\right)\right)$

Domain is the set of all input values where function is well defined and objective.

Part (a) Step 3. Explanation.

We have,

$f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$

Using definition of the composite function,

$$\begin{gathered} f\hspace{0.17em}\circ \hspace{0.17em}g=f\left(g\left(x\right)\right) \\ =f\left(\frac{2}{x}\right) \\ =\frac{3}{\frac{2}{x}-1} \\ =\frac{3}{\frac{-x+2}{x}} \\ =\frac{3x}{-x+2} \end{gathered}$$

We have given functions are the rational functions and rational functions are undefined when denominator is equal to zero.

So f(x) is undefined only when x = 1 and g(x) is undefined when x = 0.

So domain of the f(x) is set of real numbers except x = 1 and domain of the function g(x) is set of real numbers except x = 0.

So we need to exclude from the domain of g(x) that value of x for which g(x) = 1.

$$\begin{gathered} \frac{2}{x}=1 \\ \Rightarrow x=2 \end{gathered}$$

So domain of the composite function is the set of real numbers except x = 2 and x = 0.

Hence, domain of the composite function is, $(-\infty ,0) \cup (0, 2) \cup (2, \infty )$

Part (a) Step 4. Conclusion.

Hence, composite function of the function of $f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$ is $f\hspace{0.17em}\circ \hspace{0.17em}g=\frac{3x}{-x+2}$ and domain is $(-\infty ,0) \cup (0, 2) \cup (2, \infty )$.

Part (b) Step 2. Explanation.

We have,

$f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$

Using definition of the composite function,

$g\hspace{0.17em}\circ \hspace{0.17em}f=g\left(f\left(x\right)\right)$

         $$\begin{gathered} =g\left(\frac{3}{x-1}\right) \\ =\frac{2}{\frac{3}{x-1}} \\ =\frac{2\left(x-1\right)}{3} \end{gathered}$$

We have given functions are the rational functions and rational functions are undefined when denominator is equal to zero.

So $f\left(x\right)$is undefined only when x = 1 and g(x) is undefined when x = 0.

So domain of the f(x) is set of real numbers except x = 1 and domain of the function g(x) is set of real numbers except x = 0.

So we need to exclude from the domain of $f\left(x\right)$that value of x for which

$f\left(x\right)=0$.

We know that there is no such point $f\left(x\right)=0.$

So domain of the composite function is the set of real numbers except

$x=1$.

Part (b) Step 2. Conclusion.

$f\left(x\right)=\frac{3}{x-1},\hspace{0.17em}g\left(x\right)=\frac{2}{x}$is $g\hspace{0.17em}\circ \hspace{0.17em}f=\frac{2\left(x-1\right)}{3}$ and domain is