Let $h(x)=(f \circ g)(x)=\frac{1}{(x-2)^{4}}$ Find $f(x)$ given $g(x)=x-2$.
\[
f(x)=
\]
help (formulas)
Answer
So, our result is correct, and the final answer is \(\boxed{f(x)=\frac{1}{(x-4)^{4}}}\)
Step 1 ：Substitute \(g(x)\) into \(h(x)\) to get \(h(g(x))=f(g(g(x)))=\frac{1}{(g(x)-2)^{4}}\)
Step 2 ：Substitute \(x-2\) for \(g(x)\) in the equation to get \(f(g(g(x)))=\frac{1}{((x-2)-2)^{4}}\)
Step 3 ：Simplify the equation to get \(f(g(g(x)))=\frac{1}{(x-4)^{4}}\)
Step 4 ：So, the function we are looking for is \(f(x)=\frac{1}{(x-4)^{4}}\)
Step 5 ：Check our result by substituting \(g(x)\) into \(f(x)\) to see if we get \(h(x)\)
Step 6 ：We get \(f(g(x))=\frac{1}{((x-2)-4)^{4}}=\frac{1}{(x-2)^{4}}=h(x)\)
Step 7 ：So, our result is correct, and the final answer is \(\boxed{f(x)=\frac{1}{(x-4)^{4}}}\)
