Problem

2. If $f(x)=\sqrt{2-x}$ and $f(g(x))=\sqrt{2-x^{\prime}}$, then what is $g(x)$ ?

Answer

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Answer

\(\boxed{g(x) = x^\prime}\)

Steps

Step 1 ：Given that $f(x) = \sqrt{2-x}$ and $f(g(x)) = \sqrt{2-x^\prime}$, we need to find the function $g(x)$ such that $f(g(x)) = f(x^\prime)$.

Step 2 ：Substitute $g(x)$ into $f(x)$: $f(g(x)) = \sqrt{2 - g(x)}$

Step 3 ：Since $f(g(x)) = \sqrt{2 - g(x)}$ and $f(x^\prime) = \sqrt{2 - x^\prime}$, we can equate the two expressions: $\sqrt{2 - g(x)} = \sqrt{2 - x^\prime}$

Step 4 ：Comparing the two square roots, we can see that $g(x) = x^\prime$

Step 5 ：\(\boxed{g(x) = x^\prime}\)

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