Given the following function, $h ( x )$, find two functions $f$ and $g$ such that $( f \circ g ) ( x ) = h ( x )$.
$$h ( x ) = \sqrt [ 3 ] { x ^ { 2 } - 11 }$$
A) $f ( x ) = \sqrt [ 3 ] { x ^ { 2 } } , g ( x ) = - 11$
B) $f ( x ) = \sqrt [ 3 ] { x ^ { 2 } } , g ( x ) = x - 11$
C) $f ( x ) = \sqrt [ 3 ] { x } , g ( x ) = x - 11$
D) $f ( x ) = \sqrt [ 3 ] { x - 11 } , g ( x ) = x ^ { 2 }$
E) $f ( x ) = \sqrt [ 3 ] { x - 11 } , g ( x ) = x + 11$

Question

Quizplus · Accepted Answer

We want to isolate the quantity inside the cube root symbol in the expression for $h(x)$. So, let's set $x^2-11$ equal to something that we can express as a cube. If we let $u=x^2-11$, then we have that $u^{\frac{1}{3}}=\sqrt [3]{x^2-11}$. Using this, we can write:
\begin{align*}
h(x)&=\sqrt [3]{x^2-11}\
&=\sqrt [3]{u}\
&=[(\sqrt [3]{u})^2]^{\frac{1}{3}}\
&=(u^{\frac{1}{3}})^{\frac{2}{3}}\
&=(x^2-11)^{\frac{2}{3}}.
\end{align*}
So, we have found that $h(x)=(x^2-11)^{\frac{2}{3}}$. Now, we need to find two functions, $f$ and $g$, such that $(f\circ g)(x)=h(x)$. Let's look at the answer choices to see which combination of functions produces $(f\circ g)(x)=(x^2-11)^{\frac{2}{3}}$. 
\begin{align*}
&\textbf{(A)}\ f(x)=\sqrt[3]{x^2},\ g(x)=-11\
&\textbf{(B)}\ f(x)=\sqrt[3]{x^2},\ g(x)=x-11\
&\textbf{(C)}\ f(x)=\sqrt[3]{x},\ g(x)=x-11\
&\textbf{(D)}\ f(x)=\sqrt[3]{x-11},\ g(x)=x^2\
&\textbf{(E)}\ f(x)=\sqrt[3]{x-11},\ g(x)=x+11
\end{align*}
We can rule out choices (A), (B), and (C) because they have cube roots of quantities that are not present in the expression for $h(x)$. For example, in choice (A), we have $\sqrt[3]{x^2}$, which will introduce an extra square root when we square it later. In choice (B), we have $\sqrt[3]{x^2}$ again and also an extra constant term, $-11$, that doesn't appear in the expression for $h(x)$. In choice (C), we have $\sqrt[3]{x}$, which will introduce a square root of $x$ when we cube it later. Therefore, we are left with choices (D) and (E). We can check that choice (E) does not work by plugging in $g(x)=x+11$ and $f(x)=\sqrt[3]{x-11}$ into $(f\circ g)(x)$ and simplifying:
\begin{align*}
(f\circ g)(x)&=f(g(x))\
&=f(x+11)\
&=\sqrt[3]{(x+11)-11}\
&=\sqrt[3]{x}.
\end{align*}
This is not equivalent to $(x^2-11)^{\frac{2}{3}}$. Now, let's try choice (D):
\begin{align*}
(f\circ g)(x)&=f(g(x))\
&=f(x^2)\
&=\sqrt[3]{x^2-11}.
\end{align*}
This is precisely the expression for $h(x)$, so $(f\circ g)(x)$ does indeed equal $h(x)$ when we choose $f(x)=\sqrt[3]{x-11}$ and $g(x)=x^2$. Therefore, the answer is $\boxed{\textbf{(D)}}$.

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