Solve the problem.
-Let $f ( x ) = \sqrt { x - 2 }$. Find a function $y = g ( x )$ so that $( f \circ g ) ( x ) = \sqrt { x ^ { 2 } - 2 }$.
A) $g ( x ) = x ^ { 2 } + 2$
B) $g ( x ) = 2 x$
C) $g ( x ) = x ^ { 2 }$
D) $g ( x ) = x ^ { 2 } - 2$

Question

Quizplus · Accepted Answer

We need to find the function $g(x)$ such that $(f\circ g)(x) = \sqrt{x^2-2}$. Notice that $f(x)$ involves the square root of $x-2$ and we want to end up with the square root of $x^2-2$. This suggests that we need to square something in $g(x)$. In particular, notice that if we square $g(x)$, then we have:
\begin{align*}
(f\circ g)(x) &= f(g(x))\
&= \sqrt{g(x)-2}\
&= \sqrt{g(x)^2 - 2}\
&= \sqrt{(x^2-2)-2}\
&= \sqrt{x^2-2}.
\end{align*}
Thus, $g(x)$ must be of the form $g(x) = \boxed{	extbf{(C) }x^2}$, which gives us $(f\circ g)(x) = f(x^2) = \sqrt{x^2-2}$ (since $(x^2-2)-2 = x^2-4$).

Note that the other answer choices don't work because:
\begin{align*}
	extbf{(A)} &\quad (f\circ g)(x) = f(x^2+2) = \sqrt{x^2} \implies f(x^2+2)
eq \sqrt{x^2-2} \
	extbf{(B)} &\quad (f\circ g)(x) = f(2x) = \sqrt{2x-2} 
eq \sqrt{x^2-2} \
	extbf{(D)} &\quad (f\circ g)(x) = f(x^2-2) = \sqrt{x^2-4} 
eq \sqrt{x^2-2}
\end{align*}

Solve the Problem f(x)=x−2f ( x ) = \sqrt { x - 2 }f(x)=x−2​

Solve the Problem $f ( x ) = \sqrt { x - 2 }$