Question

Find the functions (a) \(f \circ g\) (b) \(g \circ f\) (c) \(f \circ f\), and (d) \(g \circ g\) and their domains

38. \(f\left( x \right) = \frac{x}{{x + {\bf{1}}}}\) and \(g\left( x \right) = {\bf{2}}x - {\bf{1}}\)

Short answer

(a) \(f \circ g\left( x \right) = 1 - \frac{1}{{2x}}\) and domain of the function is \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

(b) \(g \circ f\left( x \right) = \frac{{x - 1}}{{x + 1}}\) and domain of the function is \(\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)\).

(c) \(f \circ f\left( x \right) = \frac{x}{{2x + 1}}\) and domain of the function is \(\left( { - \infty , - 1} \right) \cup \left( { - 1, - \frac{1}{2}} \right) \cup \left( { - \frac{1}{2},\infty } \right)\).

(d) \(g \circ g\left( x \right) = 4x - 3\) and domain of the function is \(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Find the function \(f \circ g\) and its domain

Thefunction \(f \circ g\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}f \circ g\left( x \right) &= f\left( {g\left( x \right)} \right)\\ &= f\left( {2x - 1} \right)\\ &= \frac{{2x - 1}}{{2x - 1 + 1}}\\ &= 1 - \frac{1}{{2x}}\end{aligned}\)

For the function \(f \circ g\left( x \right)\) to exist, \(x \ne 0\).

So, the domain of the function is \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\).

Find the function \(g \circ f\) and its domain

The function \(g \circ f\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}g \circ f\left( x \right) &= g\left( {f\left( x \right)} \right)\\ &= g\left( {\frac{x}{{x + 1}}} \right)\\ &= 2\left( {\frac{x}{{x + 1}}} \right) - 1\\ &= \frac{{x - 1}}{{x + 1}}\end{aligned}\)

For the function \(g \circ f\left( x \right)\) to exist;

\(\begin{aligned}x + 1 &\ne 0\\x &\ne - 1\end{aligned}\)

So, thedomainof the function is \(\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)\).

Find the function \(f \circ f\) and its domain

The function \(f \circ f\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}f \circ f\left( x \right) &= f\left( {f\left( x \right)} \right)\\ &= f\left( {\frac{x}{{x + 1}}} \right)\\ &= \frac{{\left( {\frac{x}{{x + 1}}} \right)}}{{\left( {\frac{x}{{x + 1}}} \right) + 1}}\\ &= \frac{x}{{2x + 1}}\end{aligned}\)

For the function \(f \circ f\left( x \right)\) to exist;

\(\begin{aligned}2x + 1 &\ne 0\\x &\ne - \frac{1}{2}\end{aligned}\)

And,

\(\begin{aligned}x + 1 &\ne 0\\x &\ne - 1\end{aligned}\)

The domain of the function is \(\left( { - \infty , - 1} \right) \cup \left( { - 1, - \frac{1}{2}} \right) \cup \left( { - \frac{1}{2},\infty } \right)\).

Find the function \(g \circ g\) and its domain

The function \(g \circ g\left( x \right)\) can be evaluated as follows:

\(\begin{aligned}g \circ g\left( x \right) &= g\left( {g\left( x \right)} \right)\\ &= g\left( {2x - 1} \right)\\ &= 2\left( {2x - 1} \right) - 1\\ &= 4x - 3\end{aligned}\)

The domain of the function is \(\left( { - \infty ,\infty } \right)\).