Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 35

We need to find (a) $f \circ g$, (b) $g \circ f$, (c) $f \circ f$, and (d) $g \circ g$ and state their domains


$\displaystyle f(x) = x + \frac{1}{x} , \qquad \quad g(x) = \frac{x+1}{x+2} $


$
\begin{equation}
\begin{aligned}
\text{(a)} \qquad \quad f \circ g &= f(g(x)) \\

\displaystyle f \left(\frac{x+1}{x+2}\right) &= x + \frac{1}{x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle f \left(\frac{x+1}{x+2}\right) &= \frac{x+1}{x+2}+ \frac{1}{\frac{x+1}{x+2}}
&& \text{ Simplify the equation }\\

\displaystyle f \left(\frac{x+1}{x+2}\right) &= \frac{x+1}{x+2} + \frac{x+2}{x+1}
&& \text{ Get the LCD}\\

\displaystyle f \left(\frac{x+1}{x+2}\right) &= \frac{(x+1)^2+(x+2)^2}{(x+2)(x+1)}
&& \text{ Expand the equation that has an exponent}\\

\displaystyle f \left(\frac{x+1}{x+2}\right) &= \frac{x^2+2x+1+x^2+4x+4}{(x+2)(x+1)}
&& \text{ Combine like terms}

\end{aligned}
\end{equation}
$


$ \boxed{\displaystyle f \circ g = \frac{2x^2+6x+5}{(x+2)(x+1)}} $


$\boxed{ \text{ The domain of this function is } (-\infty,-2) \bigcup (-2,-1,)\bigcup (-1,\infty)} $



$
\begin{equation}
\begin{aligned}
\text{(b)} \qquad \quad g \circ f &= g(f(x)) \\

\displaystyle g\left(x+\frac{1}{x}\right) &= \frac{x+1}{x+2}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle g\left(x+\frac{1}{x}\right) &= \frac{x+\frac{1}{x}+1}{x+\frac{1}{x}+2}
&& \text{ Get the LCD of the numerator and the denominator} \\

\displaystyle g\left(x+\frac{1}{x}\right) &= \frac{\frac{x^2+1+x}{\cancel{x}}}{\frac{x^2+1+2x}{\cancel{x}}}
&& \text{ Cancel out like terms}

\end{aligned}
\end{equation}
$


$\boxed{\displaystyle g \circ f = \frac{x^2+x+1}{x^2+2x+1}} $

To get the domain of the given function, we need to get the factor of the denominator. So we have

$\displaystyle g \circ f = \frac{x^2 + x + 1}{(x + 1)^2}$

$ \boxed{ \text{ Therefore the domain of this function is }(-\infty, -1) \bigcup(-1,\infty)} $



$
\begin{equation}
\begin{aligned}
\text{(c)} \qquad \quad f \circ f &= f(f(x)) \\

\displaystyle f\left(x + \frac{1}{x}\right) &= x + \frac{1}{x}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle f\left(x + \frac{1}{x}\right) &= x + \frac{1}{x} + \frac{1}{x+ \frac{1}{x}}
&& \text{ Get the LCD of the last term, then simplify}\\

\displaystyle f\left( \displaystyle x + \frac{1}{x}\right) &= x+\frac{1}{x}+ \frac{x}{x^2+1}
&& \text{ Get the LCD of the equation} \\

\displaystyle f\left( \displaystyle x + \frac{1}{x}\right) &= \frac{x^2(x^2+1)+(x^2+1)+x^2}{x(x^2+1)}
&& \text{Simplify the equation}\\


\displaystyle f\left( \displaystyle x + \frac{1}{x}\right) & =\frac{x^4+x^2+x^2+x^2 + 1}{x^3+x}
&& \text{Combine like terms}

\end{aligned}
\end{equation}
$


$ \boxed{\displaystyle f \circ f = \frac{x^4+3x^2+1}{x^3+x}} $


$ \boxed{ \text{ The domain of this function is } (-\infty,0)\bigcup(0,\infty)} $





$
\begin{equation}
\begin{aligned}
\text{(d)} \qquad \quad g \circ g &= g(g(x)) \\

\displaystyle g \left( \displaystyle \frac{x+1}{x+2}\right) &= \displaystyle \frac{x+1}{x+2}
&& \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\

\displaystyle g \left(\displaystyle \frac{x+1}{x+2}\right) &= \displaystyle \frac{\displaystyle \frac{x+1}{x+2}+1}{\displaystyle \frac{x+1}{x+2}+2}
&& \text{Get the LCD of the numerator and the denominator}\\

\displaystyle g \left(\displaystyle \frac{x+1}{x+2}\right) &= \displaystyle \frac{\displaystyle \frac{x+1+x+2}{\cancel{x+2}}}{\displaystyle \frac{x+1+2(x+2)}{\cancel{x+2}}}
&& \text {Cancel out like terms, then simplify}

\end{aligned}
\end{equation}
$


$ \boxed{\displaystyle g \circ g = \frac{2x+3}{3x+5}} $


$\boxed{ \text{ The domain of this function is } (-\infty, \frac{-5}{3}) \bigcup (\frac{-5}{3}, \infty)}$