Algebra - Combining Functions

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Section 3.6 : Combining Functions

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4. Given \(f\left( x \right) = 6{x^2}\) and \(g\left( x \right) = {x^2} + 3x - 1\) compute each of the following.

\(\left( {g\,f} \right)\left( x \right)\)
\(\left( {f \circ g} \right)\left( x \right)\)
\(\left( {g \circ f} \right)\left( x \right)\)
\(\left( {f \circ f} \right)\left( x \right)\)

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a \(\left( {g\,f} \right)\left( x \right)\) Show Solution

Not much to do here other than do the multiplication. Also, remember that in this case the “evaluation” really only consists of plugging the two equations for each function in and doing some basic algebraic manipulations to simplify the answer if possible.

\[\left( {g\,f} \right)\left( x \right) = g\left( x \right)f\left( x \right) = \left( {{x^2} + 3x - 1} \right)\left( {6{x^2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{6{x^4} + 18{x^3} - 6{x^2}}}\]

b \(\left( {f \circ g} \right)\left( x \right)\) Show Solution

Remember that the “\( \circ \)” denotes composition and not multiplication! Also remember that the order of the composition as written in the statement needs to be followed!

\[\begin{align*}\left( {f \circ g} \right)\left( x \right) = f\left[ {g\left( x \right)} \right] & = f\left[ {{x^2} + 3x - 1} \right]\\ & = 6{\left( {{x^2} + 3x - 1} \right)^2}\\ & = 6\left( {{x^4} + 6{x^3} + 7{x^2} - 6x + 1} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{6{x^4} + 36{x^3} + 42{x^2} - 36x + 6}}\end{align*}\]

c \(\left( {g \circ f} \right)\left( x \right)\) Show Solution

Remember that the “\( \circ \)” denotes composition and not multiplication! Also remember that the order of the composition as written in the statement needs to be followed!

\[\left( {g \circ f} \right)\left( x \right) = g\left[ {f\left( x \right)} \right] = g\left[ {6{x^2}} \right] = {\left( {6{x^2}} \right)^2} + 3\left( {6{x^2}} \right) - 1 = \require{bbox} \bbox[2pt,border:1px solid black]{{36{x^4} + 18{x^2} - 1}}\]

d \(\left( {f \circ f} \right)\left( x \right)\) Show Solution

Remember that the “\( \circ \)” denotes composition and not multiplication and do not get excited about the fact that each portion is the same function. It works exactly like every other composition problem.

\[\left( {f \circ f} \right)\left( x \right) = f\left[ {f\left( x \right)} \right] = f\left[ {6{x^2}} \right] = 6{\left( {6{x^2}} \right)^2} = 6\left( {36{x^4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{216{x^4}}}\]