Introduction

Suppose that we wish to find an algebraic expression for (f \circ g)(x), where f(x) = 2x + 3 and g(x) = x - 1.

The composite function can be calculated by taking the expression for the inner function {\color{blue}g(x)} and passing it as input to the outer function f({\color{blue}x})\mathbin{:}

\begin{align*} (f \circ g)(x) &= f({\color{blue}g(x)}) \\[3pt] &= 2({\color{blue}g(x)}) + 3 \\[3pt] &= 2({\color{blue}x - 1}) + 3 \\[3pt] &= 2x - 2 + 3 \\[3pt] &= 2x + 1 \end{align*}

So, we just take the expression for f and replace every occurrence of x with the expression for g(x).

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Example: Finding a Composite Function

Find (f \circ g)(x) for f(x) = 3x^2 - 2x and g(x) = x + 4.

EXPLANATION

Note that (f \circ g)(x) is equivalent to f(g(x)).

To find f(g(x)), we write down the function f and replace every occurrence of x with g(x)\mathbin{:}

\begin{align} f(x) &= 3{\color{blue}x}^2-2{\color{blue}x} \\[3pt] f(g(x)) &= 3({\color{blue}g(x)})^2 - 2({\color{blue}g(x)}) \end{align}

Then, we substitute {\color{blue}g(x)} = {\color{red}x+4} into the right-hand side of the above and simplify, as follows:

\begin{align*} f(g(x)) &= 3({\color{blue}g(x)})^2 - 2({\color{blue}g(x)}) \\[3pt] &= 3({\color{red}x + 4})^2 - 2({\color{red}x + 4}) \\[3pt] &= 3(x + 4)(x+4) - 2x - 8 \\[3pt] &= 3(x^2 +4x + 4x + 16) - 2x - 8 \\[3pt] &= 3(x^2 +8x + 16) - 2x - 8 \\[3pt] &= 3x^2 + 24x + 48 - 2x - 8 \\[3pt] &= 3x^2 + 22x + 40 \end{align*}

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Practice: Finding a Composite Function

What is $f(g(x))$ for $f(x) = x + 3$ and $g(x) = 2x + 1?$

a
 $2x - 7$
b
 $2x + 7$
c
 $2x + 4$
d
 $2x - 4$
e
 $2x + 1$
Practice: Finding a Composite Function

If $f(x) = x + 5$ and $g(x) = 3x + 1$, then $(g \circ f)(x) = $

a
 $3x - 16$
b
 $3x + 6$
c
 $3x + 1$
d
 $3x + 16$
e
 $3x + 5$
Practice: Finding a Composite Function

If $f(x) = x + 1$ and $g(x) = x^2 - 2x$, then $(g \circ f)(x) = $

a
b
c
d
e
Example: Composing a Function With Itself

If g(x)=x^2+x, then find (g\circ g)(x).

EXPLANATION

First, we note that (g\circ g)(x) is equivalent to g(g(x)).

To find g(g(x)), we write down the function g and replace every occurrence of x with g(x)\mathbin{:}

\begin{align} g(x) &= {\color{blue}x}^2+{\color{blue}x} \\[3pt] g(g(x)) &= ({\color{blue}g(x)})^2 +{\color{blue}g(x)} \end{align}

Then, we substitute {\color{blue}g(x)} = {\color{red}x^2+x} into the right-hand side of the above and simplify, as follows:

\begin{align} g(g(x)) &=({\color{blue}g(x)})^2+{\color{blue}g(x)} \\[3pt] &=({\color{red}x^2+x})^2+({\color{red}x^2+x}) \\[3pt] &=(x^2+x)(x^2+x)+x^2 + x \\[3pt] &=(x^4+x^3+x^3+x^2) + x^2+x \\[3pt] &=x^4+2x^3+2x^2+x \end{align}

Therefore, g(g(x)) = x^4+2x^3+2x^2+x.

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Practice: Composing a Function With Itself

If $g(x)=-2x+5,$ then $(g\circ g)(x)=$

a
 $4x+5$
b
 $4x+15$
c
 $4x-5$
d
 $-4x-5$
e
 $-4x+5$
Practice: Composing a Function With Itself

If $g(x) = x^2+2,$ then $(g\circ g)(x) = $

a
b
c
d
e
Example: Evaluating a Function With a Variable Input

If f(x) = x^2-1 , then find f(a+1).

EXPLANATION

To find f(a+1), we write down the function f and replace every occurrence of x with (a+1)\mathbin{:}

\begin{align} f({\color{blue}x}) &= {\color{blue}x}^2-1\\[3pt] f({\color{blue}a+1}) &= ({\color{blue}a+1})^2 - 1 \end{align}

Then, we simplify as follows:

\begin{align*} f(a + 1) &= (a+1)^2 - 1 \\[3pt] &= (a+1)(a+1) - 1 \\[3pt] &= a^2 + a + a + 1 - 1 \\[3pt] &= a^2 + 2a \end{align*}

Therefore, f(a+1)=a^2+2a.

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Practice: Evaluating a Function With a Variable Input

If $f(x) = x^2 - 2x - 1$, then find an expression for $f(a+1).$

a
b
c
d
e
Practice: Evaluating a Function With a Variable Input

If $f(x) = x^2 - 2x - 1$, then find an expression for $f(a+b).$

a
b
c
d
e
Example: Solving Equations Involving Composite Function

If f(x) = r + 4x and f(5a+1) = 22a, where a and r are constants, find an expression for r in terms of a.

EXPLANATION

We are given the linear function f(x) = r + 4x, and are told that f(5a+1) = 22a.

To find f(5a+1) in terms of r, we write down the function f and replace every occurence of x with 5a+1{:}

\begin{align*} f({\color{blue}x}) &= r + 4{\color{blue}x} \\[5pt] f({\color{blue}5a+1}) &= r + 4({\color{blue}5a+1}) \end{align*}

Therefore, we simplify f(5a+1) = 22a as follows:

\begin{align*} f({\color{blue}5a+1}) &= 22a \\[5pt] r + 4({\color{blue}5a+1}) &= 22a \\[5pt] r + 20a+4 &= 22a \end{align*}

Finally, we solve for r{:}

\begin{align*} r + 20a + 4 &= 22a \\[5pt] r &= 22a - 20a - 4 \\[5pt] r &= 2a-4 \end{align*}

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Practice: Solving Equations Involving Composite Function

If $f(x) = b + 5x$ and $f(3a+2) = 20a,$ where $a$ and $b$ are constants, find an expression for $b$ in terms of $a.$

a
 $5a-10$
b
 $3a+10$
c
 $5b-10$
d
 $5a-12$
e
 $4a-5$
Practice: Solving Equations Involving Composite Function

If $f(x) = b - 5x$ and $f(3a+2) = 20a,$ where $a$ and $b$ are constants, find an expression for $b$ in terms of $a.$

a
b
c
d
e
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