Composition of Functions

Solved Problems

Solution.

Draw the arrow diagram for the composition of functions \(f \circ g:\)

It follows from the diagram:

Hence, the composite function \(f \circ g\) is given by

The domain of \(f \circ g\) is the set \(\left\{ {5,7,9,2} \right\},\) and the range is the set \(\left\{ {7,1} \right\}.\)

Solution.

By definition, \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).\)

The inner function \({g\left( x \right)}\) is defined for all \(x\) except \(x = \frac{5}{4}.\) Hence, the point \(x = \frac{5}{4}\) must be excluded from the domain of the composition \(f \circ g.\)

The outer function \({f\left( x \right)}\) is defined for all \(x\) except \(x = \frac{4}{3}.\) So, we also need to exclude those values of \(x\) for which the inner function \({g\left( x \right)}\) is equal to \(\frac{4}{3}.\) Determine these values.

Thus, the domain of \(f \circ g\) consists of all real values of \(x\) except the points \(x = \frac{5}{4}, \frac{29}{16}.\) This can be written in the form

Solution.

1. The composite function \(\left(f \circ g\right)\left(x\right)\) is written as \(f\left( {g\left( x \right)} \right).\) We take the outer function \(f\left( x \right)\) and substitute the inner function \({g\left( x \right)}\) for \(x:\)
   \[\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) = {g^2}\left( x \right) + 2g\left( x \right) = {\left( {{x^2} - x} \right)^2} + 2\left( {{x^2} - x} \right) = \color{magenta}{x^4} - \color{green}{2{x^3}} + \color{red}{x^2} + \color{red}{2{x^2}} -\color{blue}{2x} = \color{magenta}{x^4} -\color{green}{2{x^3}} + \color{red}{3{x^2}} - \color{blue}{2x}.\]
2. Calculate the composition \(g \circ f:\)
   \[\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) = {f^2}\left( x \right) - f\left( x \right) = {\left( {{x^2} + 2x} \right)^2} - \left( {{x^2} + 2x} \right) = \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} - \color{red}{x^2} - \color{blue}{2x} = \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{3{x^2}} - \color{blue}{2x}.\]
   Notice that \(f \circ g \ne g \circ f,\) that is the composition of functions is not commutative.
3. The composition of functions \(f \circ f\) is given by
   \[\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) = {f^2}\left( x \right) + 2f\left( x \right) = {\left( {{x^2} + 2x} \right)^2} + 2\left( {{x^2} + 2x} \right) = \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} + \color{red}{2{x^2}} + \color{blue}{4x} = \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{6{x^2}} + \color{blue}{4x}.\]
4. Similarly we determine the function \(g \circ g:\)
   \[\require{cancel}\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) {g^2}\left( x \right) - g\left( x \right) = {\left( {{x^2} - x} \right)^2} - \left( {{x^2} - x} \right) = \color{magenta}{x^4} - \color{green}{2{x^3}} + \cancel{\color{red}{x^2}} - \cancel{\color{red}{x^2}} + \color{blue}{x} = \color{magenta}{x^4} - \color{green}{2{x^3}} + \color{blue}{x}.\]

Solution.

1. The find the composite function \(f \circ g \circ f,\) we first determine the composition of inner functions \(g \circ f:\)
   \[\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) = g\left( {{x^2} + 1} \right) = \sqrt {{x^2} + 1} .\]
   Then, using the associative law, the triple composition is given by
   \[\left(f \circ g \circ f\right)\left(x\right) = \left(f \circ \left({g \circ f}\right)\right)\left(x\right) = f\left[ {g\left( {f\left( x \right)} \right)} \right] = f\left( {\sqrt {{x^2} + 1} } \right) = {\left( {\sqrt {{x^2} + 1} } \right)^2} + 1 = {x^2} + 1 + 1 = {x^2} + 2.\]
2. Here we first find the composition \(f \circ g:\)
   \[\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) = f\left( {\sqrt x } \right) = {\left( {\sqrt x } \right)^2} + 1 = x + 1.\]
   Hence,
   \[\left(g \circ f \circ g\right)\left(x\right) = \left(g \circ \left( {f \circ g} \right)\right)\left(x\right) = g\left[ {f\left( {g\left( x \right)} \right)} \right] = g\left( {x + 1} \right) = \sqrt {x + 1} .\]
3. Calculate the composition of functions \(g \circ f \circ f\) using the same approach. Since
   \[\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) = f\left( {{x^2} + 1} \right) = {\left( {{x^2} + 1} \right)^2} + 1 = {x^4} + 2{x^2} + 1 + 1 = {x^4} + 2{x^2} + 2,\]
   the triple composition \(g \circ f \circ f\) is given by
   \[\left(g \circ f \circ f\right)\left(x\right) = \left(g \circ \left( {f \circ f} \right)\right)\left(x\right) = g\left[ {f\left( {f\left( x \right)} \right)} \right] = g\left( {{x^4} + 2{x^2} + 2} \right) = \sqrt {{x^4} + 2{x^2} + 2} .\]

Solution.

Find the inverse function \(f^{-1}\) by solving the equation \(y = \sqrt {{x^3} + 1}\) for \(x:\)

To calculate the composition of functions \(f^{-1} \circ f,\) we substitute \(y = \sqrt {{x^3} + 1}\) in the equation \(x = \sqrt[3]{{{y^2} - 1}}.\) For any \(x\) in the domain of \(f,\) we have

Thus, \({f^{ - 1}} \circ f = I,\) where \(I\) is the identity function in the domain of \(f.\)

Solution.

Calculate the compositions of functions \(f \circ g\) and \(g \circ f:\)

Plug both expressions into the original equation and solve it for \(x:\)