Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 38

We need to find the function $f \circ g \circ h$

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


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

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

g(h(x)) =& x^2\\

g( 1 - x) =& x^2
&& \text{ Substitute the given function $h(x)$ to the value of $x$ of the function $g(x)$}\\

g(1 - x) =& (1 -x)^2
&& \text{ Using FOIL method}\\

g \circ h =& 1 - 2x + x^2\\
\\
\text{ Solving for $f \circ g \circ h$}\\

g \circ h =& 1 - 2x + x^2\\

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

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

f(1 - 2x + x^2) =& 2(1 -2x + x^2) -1
&& \text{ Simplify the equation}\\

f(1 -2x + x^2) =& 2 - 4x + 2x^2 -1
&& \text{ Combine like terms}


\end{aligned}
\end{equation}
$


$\boxed{f \circ g \circ h = 1 - 4x + 2x^2}$

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