Question

For the given functions f and g, find:

$a)f\circ gb)g\circ fc)f\circ fd)g\circ g$

State the domain of each composite function.

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

Short answer

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

Step-by-step solution

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.

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

Domainis 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 thef(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