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Section 1.10 Combinations of Functions
Definition 1.10.1 .
The function \(f \circ g\text{,}\) read as \(f\) of \(g\) is a composition of \(f(x)\) and \(g(x)\) where
\begin{equation*}
(f\circ g)(x)=f(g(x))
\end{equation*}
here, all \(x\) in the domain of \(g(x)\) is also in the domain of \(f(x)\text{.}\)
If
\(f \circ g (x) = g \circ f (x) \) then
\(f(x)\) is inverse function of
\(g(x).\)
Example 1.10.3 .
If
\(f(x) = x^2 \) and
\(g(x) = x-5 \) then find the composite functions
\(f\circ g\) and
\(g\circ f.\)
Solution .
\begin{equation*}
(f\circ g)(x) = f(g(x)) = (g(x))^2 = (x-5)^2
\end{equation*}
This is the value of function \(f\) at \(g(x)\text{.}\) Hence, replace \(x\) of \(f(x)\) with \(g(x).\)
\begin{equation*}
(g\circ f)(x) = g(f(x)) = (f(x) -5) = (x^2-5)
\end{equation*}
This is the value of function \(g\) at \(f(x)\text{.}\) Hence, replace \(x\) of \(g(x)\) with \(f(x).\)
Example 1.10.4 .
If \(f(x) =\sin x\) and \(g(x) = x^2,\) then find
\(\displaystyle (f\circ g)(x) =f(g(x)) \)
\(\displaystyle (g\circ f)(x) =g(f(x)) \)
Solution .
\(\displaystyle (f\circ g)(x) =f(g(x)) = \sin (g(x)) = \sin (x^2)\)
\(\displaystyle (g\circ f)(x) =g(f(x)) = (f(x))^2 =(\sin x)^2 \)
Example 1.10.5 .
If
\(f(x) =x^2\) and
\(g(x) = 2^x\text{,}\) then find the value of
\(x\) for which
\(f\circ g (x) =g \circ f (x).\)
Solution .
\begin{align*}
f\circ g (x) \amp = g \circ f (x) \\
f(g(x)) \amp =g(f(x)\\
(g(x))^2 \amp = 2^{f(x)}\\
(2^{x})^2 \amp =2^{x^2} \\
2^{2x} \amp = 2^{x^2}\\
\therefore \, 2x \amp =x^2 \\
or, \, x \amp = {0,2}.
\end{align*}
Example 1.10.6 .
Let
\(f(x) = \sqrt{x}\) \(\rightarrow \, domain:\, [0, \infty)\)
and
\(g(x) = x - 2\) \(\rightarrow \, domain:\, [-\infty, \infty)\)
Sum:
\begin{equation*}
(f + g)(x) = \sqrt{x} + x - 2
\end{equation*}
Domain = domain of \(f\) \(\cap\) domain of \(g = [0,\infty)\text{.}\)
Product:
\begin{equation*}
(f \cdot g)(x) = \sqrt{x}(x - 2)
\end{equation*}
Domain = \([0,\infty)\)
Quotient:
\begin{equation*}
\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x-2}
\end{equation*}
Domain = \([0, \infty)\) excluding where denominator = 0 at \(x - 2 = 0 \, \Rightarrow \, x=2.\) So domain = \([0,2)\cup (2,\infty)\)
Composition:
\begin{equation*}
(f\circ g)(x)=f(g(x))= \sqrt{x-2}
\end{equation*}
First apply
\((g): x - 2\text{.}\) Then apply
\(( f ):\) inside square root must be
\(\geq 0\)
\(x - 2 \geq 0\) \(\rightarrow \, x \geq 2\text{.}\) Domain
\(= [2,\infty)\)
Example 1.10.7 .
If
\(f(x)=x^2\) and
\(g(x) =\frac{1}{x}\) find
\(f\circ g\)
Solution .
By definition:
\begin{equation*}
(f\circ g)(x)=f(g(x))
\end{equation*}
First apply \(g\text{,}\) then plug the result into \(f\text{:}\)
\begin{align*}
g(x) \amp =\frac{1}{x} \\
(f\circ g)(x) \amp =f(g(x)) = f(\frac{1}{x}) = \left( \frac{1}{x} \right)^2= \frac{1}{x^2}.
\end{align*}
Example 1.10.8 .
If
\(f(x)=\sqrt{x}\text{,}\) \(g(x)=x^2+1\) then find (a)
\(f\circ g\) and (b)
\(f + g\text{.}\)
If
\(f(x)=\frac{1}{x-3}\text{,}\) and
\(g(x)=x^2\) find
\(\frac{f}{g}\text{.}\)
Answer .
\(\frac{f}{g} = \frac{f(x)}{g(x)} = \frac{1}{(x^2)(x-3)}\)
