Q 29. - 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.

$f\left(x\right)=2x+3 and g\left(x\right)=3x$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right) \hspace{0.17em}f\hspace{0.17em}\circ \hspace{0.17em}g=6x+3 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{b}\right) g\hspace{0.17em}\circ \hspace{0.17em}f\hspace{0.17em}=6x+9 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{c}\right) f\circ \hspace{0.17em}f\hspace{0.17em}=4x+9 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{d}\right) g\hspace{0.17em}\circ \hspace{0.17em}g=9x and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \end{gathered}$$

Part (a) Step 1. Given Information.

We have given functions are,

$f\left(x\right)=2x+3 and g\left(x\right)=3x$

Part (a) Step 2. Concept used.

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

$(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$

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

Part (a) Step 3. Explanation.

We have given,

$f\left(x\right)=2x+3 and g\left(x\right)=3x$

Using definition of composite function,

$$\begin{gathered} (f\circ g)\left(x\right)=f\left(g\right(x\left)\right) \\ =f\left(3x\right) \\ =2\left(3x\right)+3 \\ =6x+3 \end{gathered}$$

We have given functions f and g both are polynomial functions so the domain of both functions is set of real numbers.

Therefore domain of the composite function is also the set of real numbers.

Domain: $(-\infty , \infty )$

Part (a) Step 4. Conclusion.

Hence, composite function of the functions $f\left(x\right)=2x+3 and g\left(x\right)=3x$ is $f\circ g=6x+3$ and domain of the composite function is $(-\infty , \infty )$.

Part (b) Step 1. Explanation.

We have,

$f\left(x\right)=2x+3 and g\left(x\right)=3x$

Using definition of the composite function,

$$\begin{gathered} (g\circ f)\left(x\right)=g\left(f\right(x\left)\right) \\ =g(2x+3) \\ =3(2x+3) \\ =6x+9 \end{gathered}$$

We have given functions are polynomial functions and domain of the polynomial functions is the set of all real numbers.

So domain of the composite function is also the set of real numbers.

That is, $(-\infty , \infty )$.

Part (b) Step 2. Conclusion.

Hence, composite function of the functions $f\left(x\right)=2x+3 and g\left(x\right)=3x$ is $g\circ f=6x+9$ and domain of the composite function is $(-\infty ,\infty )$.

Part (c) Step 1. Explanation.

We have given,

$f\left(x\right)=2x+3 and g\left(x\right)=3x$

Using definition of the composite function,

$$\begin{gathered} (f\circ f)\left(x\right)=f\left(f\right(x\left)\right) \\ =f(2x+3) \\ =2(2x+3)+3 \\ =4x+9 \end{gathered}$$

We have given functions are polynomial functions and domain of the polynomial functions is the set of all real numbers.

So domain of the composite function is also the set of real numbers.

That is, $\left(-\infty ,\hspace{0.17em}\infty \right)\hspace{0.17em}$

Part (c) Step 2. Conclusion.

Hence, composite function of $f\left(x\right)=2x+3$ with itself is $f\hspace{0.17em}\circ \hspace{0.17em}f=4x+9$

and domain of the composite function is $\left(-\infty ,\hspace{0.17em}\infty \right)$.

Part (d) Step 1. Explanation.

We have,

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

Using definition of composite function,

$$\begin{gathered} (g\circ g)\left(x\right)=g\left(g\right(x\left)\right) \\ =g\left(3x\right) \\ =3\left(3x\right) \\ =9x \end{gathered}$$

We have given functions are polynomial functions and domain of the polynomial functions is the set of all real numbers.

So domain of the composite function is also the set of real numbers.

That is, $\left(-\infty ,\hspace{0.17em}\infty \right)$

Part (d) Step 2. Conclusion.

Hence, composite function of $g\left(x\right)=3x$ with itself is $\hspace{0.17em}g\hspace{0.17em}\circ \hspace{0.17em}g=9x$ and domain of the composite function is $\left(-\infty ,\hspace{0.17em}\infty \right)$.