For the function f(x)=x^3-2x^2+2x, what is (f^-1)'(4)?

Question

A. 1/4
B. 1/6
C. 1/34
D. 6
E. 34

Answer 1

f(x)=x^3-2x^2+2x
y = x^3-2x^2+2x

the inverse is:
x = y^3 - 2y^2 + 2y

take the derivative , that's what f^-1 ' (x) means

1 = 3y^2 dy/dx - 4y dy/dx + 2dy/dx
so when y = 4
1 = 3(16) dy/dx - 16 dy/dx + 2 dy/dx
1 = 34 dy/dx
dy/dx = 1/34

Answer 2

@mathhelper

the way you did makes sense but here's how i did it and ended up getting a different ans:

since (f^-1)'(4), you know f(x)=4, so:
x^3-2x+2x=4
x=2

since (f^-1)'(x)=1/f'(f^-1(x)):
we can find the derivative of f(x):
f'(x)=3x^2+4x+2

solve for x=2:
f'(x)=3x^2+4x+2
f'(x)=6

since (f^-1)'(x)=1/f'(f^-1(x)):
(f^-1)'(x)=1/6

both of these methods seem right to me, but im not sure which answer is correct one?

Answer 3

Recall that if g(x) is the inverse of f(x) then
if f(a) = b then g'(b) = 1/f'(a)
f(2) = 4
so g'(4) = 1/f'(2) = 1/6