(f^-1)'(a) f(x)= x^3 + 7x - 1, a = -9

Answer 1

Hi,

(f^-1)'(a) f(x)= x^3 + 7x - 1, a = -9

a=-9 is input for inverse so it is out put for function

it means you need to find x when y = -9 =====> f(x)= x^3 + 7x - 1= -9

x^3 + 7x - 1 +9 = 0 you need to solve this equation that x will be -1

IT MEANS: point (-1,-9) of function is (-9,-1) of inverse

NOW [f^-1(-9)]' = 1/ f'(-1)

f'(x) = 3x²+7 ===> f'(-1) = 10 =====> [f^-1(-9)]' = 1/10

I hope it is useful,

Minoo