Q.31 - Precalculus Enhanced with Graphing Utilities

Question

For the given functions f and g, find:   

$a) f\hspace{0.17em}\circ \hspace{0.17em}\hspace{0.17em}g\hspace{0.17em} b)\hspace{0.17em}\hspace{0.17em}g\hspace{0.17em}\circ \hspace{0.17em}\hspace{0.17em}g c) f\hspace{0.17em}\circ \hspace{0.17em}f\hspace{0.17em} d) g\hspace{0.17em}\circ \hspace{0.17em}\hspace{0.17em}f$

State the domain of each composite function. 

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

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right) f\hspace{0.17em}\circ \hspace{0.17em}g=3x^2+1 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{b}\right) g\hspace{0.17em}\circ \hspace{0.17em}f= 9x^2+6x+1 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{c}\right) f\hspace{0.17em}\circ \hspace{0.17em}f=9x+4 and domain is \left(-\infty ,\hspace{0.17em}\infty \right) \\ \mathrm{Part} \left(\mathrm{d}\right) g\hspace{0.17em}\circ \hspace{0.17em}g=x^4 and domain is \left(-\infty ,\hspace{0.17em}\infty \right). \end{gathered}$$ 

Part (a) Step 1. Given Information.

We have given,

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

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, 

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

Using definition of the composite function, 

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

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: $\left(-\infty ,\infty \right)$

Part (a) Step 4. Conclusion.

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

Part (b) Step 1. Explanation.

We have given,

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

Using definition of the composite function,

$$\begin{gathered} g\hspace{0.17em}\circ \hspace{0.17em}f=g\left(f\left(x\right)\right) \\ =g\left(3x+1\right) \\ ={(3x+1)}^2 \\ =9x^2+6x+1 \end{gathered}$$

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

Part (b) Step 2. Conclusion.

Hence, composite function of the functions $f\left(x\right)=3x+1,\hspace{0.17em}g\left(x\right)=x^2$ is $g\hspace{0.17em}\circ \hspace{0.17em}f= 9x^2+6x+1$ and domain of the function is, $\left(-\infty ,\hspace{0.17em}\infty \right)$.

Part (c) Step 1. Explanation.

We have given,

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

Using definition of the composite function, 

$$\begin{gathered} f\hspace{0.17em}\circ \hspace{0.17em}f=f\left(f\left(x\right)\right) \\ =f(3x+1) \\ =3(3x+1)+1 \\ =9x+4 \end{gathered}$$

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

Part (c) Step 2. Conclusion.

Hence, composite function of the function $f\left(x\right)=3x+1,\hspace{0.17em}g\left(x\right)=x^2, is f\hspace{0.17em}\circ \hspace{0.17em}f=9x+4$ and domain is $\left(-\infty ,\hspace{0.17em}\infty \right)$.

Part (d) Step 1. Explanation.

We have,

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

Using definition of the composite function,

$$\begin{gathered} g\hspace{0.17em}\circ \hspace{0.17em}g=g\left(g\left(x\right)\right) \\ =g\left(x^2\right) \\ =x^4 \end{gathered}$$

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

Part (d) Step 2. Conclusion.

Hence, composite function of $f\left(x\right)=3x+1,\hspace{0.17em}g\left(x\right)=x^2 is g\hspace{0.17em}\circ \hspace{0.17em}g=x^4$ and domain is $\left(-\infty ,\hspace{0.17em}\infty \right)$.