Sunday, April 1, 2012

Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 39

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

$f(x) = \qquad \quad \sqrt{x - 3}, \qquad \quad g(x)=x^2, \qquad \quad h(x)= x^3 + 2$


$
\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(x^3 + 2) =& x^2
&& \\

g(x^3 + 2) =& (x^3 + 2)^2
&& \text{ Using FOIL method}\\

g \circ h =& x^6 + 4x^3 + 4\\

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

g \circ h =& x^6 + 4x^3 + 4\\

f(g(h(x))) =& \sqrt{x - 3}\\

f (x^6 + 4x^3 + 4) =& \sqrt{x - 3}
&& \text{ Substitute the value of $x$}\\

f(x^6 + 4x^3 + 4) =& \sqrt{x^6 + 4x^3 + 4 -3}
&& \text{ Combine like terms}

\end{aligned}
\end{equation}
$



$\boxed{f \circ g \circ h = \sqrt{x^6 + 4x^3 + 1}}$

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