Composite Functions
A composite function is formed from two functions in the following way.
$$(g \circ f)(x) = g(f(x))$$
If $f(x)=x+3$ and $g(x)=2x$ are two functions, then we combine the two functions to form the composite function:
\( \begin{align}
(g \circ f)(x) &= g(f(x)) \\
&= 2f(x) \\
&= 2(x+3) \\
&= 2x+6
\end{align} \)
That is, $f(x)$ replaces $x$ in the function $g(x)$.
The composite function reads $g$ of $f$ and can be written $g \circ f$.
$$(f \circ g)(x) = f(g(x))$$
Another composite function is:
\( \begin{align}
(f \circ g)(x) &= f(g(x)) \\
&= g(x) + 3 \\
&= 2x+3
\end{align} \)
In this case, $g(x)$ replaces $x$ in $f(x)$. This composite reads $f$ of $g$ and can be written $f \circ g$.
For the composite function $f(g(x))$ to be defined, the range of $g$ must be a subset of (or equal to) the domain of $f$. It is easiest first to list the domain and function of both $f(x)$ and $g(x)$ when dealing with composite function problems.
Example 1
Given $f(x)=2x-1$ and $g(x)=3x+5$, find the simplest expression of $(f \circ g)(x)$.
\( \begin{align} \displaystyle
(f \circ g)(x) &= f(g(x)) \\
&= 2g(x)-1 \\
&= 2(3x+5)-1 \\
&= 6x+10-1 \\
&= 6x+9
\end{align} \)
Example 2
Given $f(x)=4x-5$, find the simplest expression of $(f \circ f)(x)$.
\( \begin{align} \displaystyle
(f \circ f)(x) &= f(f(x)) \\
&= 4f(x)-5 \\
&= 4(4x-5)-5 \\
&= 16x-20-5 \\
&= 16x-25
\end{align} \)
Example 3
Given $f(x)=2x+7$ and $g(x)=-x+2$, find the simplest expression of $(g \circ f)(2)$.
\( \begin{align} \displaystyle
(g \circ f)(2) &= g(f(2)) \\
&= -f(2) + 2 \\
&= -(2 \times 2+7)+2 \\
&= -11 +2 \\
&= -9
\end{align} \)
Example 4
Given $f(x)=\sqrt{x}$ and $g(x)=x+1$, find the domain and range of $(g \circ f)(x)$.
\( \begin{array}{|c|c|c|}
\hline & \textit{Domain} & \textit{Range} \\ \hline
f(x)=\sqrt{x} & x \ge 0 & y \ge 0 \\ \hline
g(x)=x+1 & x \in \mathbb{R} & y \in \mathbb{R} \\ \hline
(g \circ f)(x) & x \ge 0 & y \ge 0 \\ \hline
\end{array} \)
Example 5
Given $f(x)=-2x+3$, $g(x)=5x+k$ and $(g \circ f)(x)=(f \circ g)(x)$, find $k$.
\( \begin{align} \displaystyle
(g \circ f)(x) &= (f \circ g)(x) \\
g(f(x)) &= f(g(x)) \\
5f(x)+k &= -2g(x)+3 \\
5(-2x+3)+k &= -2(5x+k)+3 \\
-10x+15+k &= -10x-2k+3 \\
3k &= 3-15 \\
\therefore k &= -4
\end{align} \)
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