4.7: Composite functions

Last updated
Save as PDF

Page ID: 40920

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Similar to the way in which we used transformations to analyze the equation of a function, it is sometimes helpful to consider a given function as being several functions of the variable combined together.

For example, instead of thinking of the function \(f(x)=(2 x-7)^{3}\) as being a single function, we can think of it as being two functions:
\[
\begin{array}{c}
g(x)=2 x-7 \\
\text { and } \\
h(x)=x^{3}
\end{array}
\]
Then \(f(x)\) is the combination or "composition" of these two functions together. The first function multiplies the variable by \(2,\) and subtracts 7 from the result. The second function takes this answer and raises it to the third power. The notation for the composition of functions is an open circle: 0

In the example above we would say that the function \(f(x)=(2 x-7)^{3}\) is equivalent to the composition \(h \circ g(x)\) or \(h(g(x))\). The order of function composition is important. The function \(g \circ h(x)\) would be equivalent to \(g(h(x)),\) which would be
\[
\text { equal to } g\left(x^{3}\right)=2\left(x^{3}\right)-7=2 x^{3}-7
\]

Exercises 4.7
Find \(f \circ g(x)\) and \(g \circ f(x)\) for each of the following problems.
1) \(\quad f(x)=x^{2} \\ g(x)=x-1\)
2) \(\quad f(x)=|x-3| \\ g(x)=2 x+3\)

3) \(\quad f(x)=\frac{x}{x-2} \\ g(x)=\frac{x+3}{x} \)
4) \(\quad f(x)=x^{3}-1 \\ g(x)=\frac{1}{x^{3}+1\)

5) \(\quad f(x)=\sqrt{x+1} \\ g(x)=x^{4}-1\)
6) \(\quad f(x)=2 x^{3}-1 \\ g(x)=\sqrt[3]{\frac{x+1}{2}\)

Find functions \(f(x)\) and \(g(x)\) so that the given function \(h(x)=f \circ g(x)\)
7) \(\quad h(x)=(3 x+1)^{2}\)
8) \(\quad h(x)=\left(x^{2}-2 x\right)^{3}\)
9) \(\quad h(x)=\sqrt{1-4 x}\)
10) \(\quad h(x)=\sqrt[3]{x^{2}-1}\)
11) \(\quad h(x)=\left(\frac{x+1}{x-1}\right)^{2}\)
12) \(\quad h(x)=\left(\frac{1-2 x}{1+2 x}\right)^{3}\)
13) \(\quad h(x)=\left(3 x^{2}-1\right)^{-3}\)
14) \(\quad h(x)=\left(1+\frac{1}{x}\right)^{-2}\)
15) \(\quad h(x)=\sqrt{\frac{x}{x-1}}\)
16) \(\quad h(x)=\sqrt[3]{\frac{x-1}{x}}\)
17) \(\quad h(x)=\sqrt{\left(x^{2}-x-1\right)^{3}}\)
18) \(\quad h(x)=\sqrt[3]{\left(1-x^{4}\right)^{2}}\)
19) \(\quad h(x)=\frac{2}{\sqrt{4-x^{2}}}\)
20) \(\quad h(x)=-\left(\frac{3}{x-1}\right)^{5}\)

21) A spherical weather balloon is inflated so that the radius at time \(t\) is given by the equation:
\[
r=f(t)=\frac{1}{2} t+2
\]
Assume that \(r\) is in meters and \(t\) is in seconds, with \(t=0\) corresponding to the time the balloon begins to be inflated. If the volume of a sphere is given by the formula:
\[
v(r)=\frac{4}{3} \pi r^{3}
\]
Find \(V(f(t))\) and use this to compute the time at which the volume of the balloon is \(36 \pi \mathrm{m}^{3}\)