let g(x) be the inverse of an invertible function f(x) which is derivatable at x=3. If f(3)=9 and f'(3)=9, write the calue of g'(9).

GIven f(3) = 9 and f '(3)=9

we have an identity bigl(f^{-1}bigr)'(b) = frac{1}{f'(a)}  where b = f(a)

here a = 3 , b = f(3) = 9

⇒g'(9) = 1/f '(3)

⇒g'(9) = 1/9

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