I tried using the inverse function theorem, but the answer I got through that was $-1$.
The answer key says the answer is $-1/2$.
Where am I going wrong?
2026-03-26 13:02:45.1774530165
If $f(1)=3$, $f'(1)=2$, $f''(1)=4$, then $(f^{-1})''(3) =$
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Let $g(x)=f^{-1}(x)$
$$ f(g(x))=x\Rightarrow f'(g(x))g'(x)=1\Rightarrow g'(x)={1\over f'(g(x))} $$
$$ g''(x)=-{f''(g(x))g'(x)\over (f'(g(x)))^2}=-{f''(g(x))\over (f'(g(x)))^3} $$
Substituting values
$$ g''(3)=-{f''(g(3))\over (f'(g(3)))^3}=-{f''(1)\over (f'(1))^3}=-{4\over8}=-{1\over2} $$