I want to find the inverse of $f(x)=\frac{3}{4}x^2-\frac{1}{4}x^3 $ when $0<x<2$.
According to wolfram the answer is inverse
I would like to know how can I find wolfram's inverse.
The correct formula is $$(f^{-1})'(y) = \dfrac{1}{f'(f^{-1}(y))}$$
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The correct formula is $$(f^{-1})'(y) = \dfrac{1}{f'(f^{-1}(y))}$$