Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 37

Find the function $f \circ g \circ h$.



$f(x) = x+1 , \qquad \quad g(x) = 2x ,\qquad \quad h(x) = x - 1$



$
\begin{equation}
\begin{aligned}
f \circ g \circ h &= f(g(h(x)))\\

\text{Solving for $g \circ h$}\\

g(h(x)) =& 2x
&& \text{Substitute the given function $h(x)$ to the value of $x$ of the function $g(x)$}\\

g(x -1) =& 2x \\

g(x -1) =& 2(x - 1)
&&\text{ Simplify the equation}\\

g \circ h =& 2x -2 \\


\text{Solving for $f \circ g \circ h$}\\

g \circ h =& 2x -2\\

f \circ g \circ h =& f(g(h(x)))\\

f(2x - 2) =& x+1
&& \text{ Substitute the value of $x$}\\

f(2x - 2) =& 2x - 2 +1
&& \text{ Combine like terms}\\

\end{aligned}
\end{equation}
$



$\qquad \qquad \boxed{f \circ g \circ h = 2x - 1}$