# Given: F(2)=3, F'(2)=4, F(3)=8, F'(3)=7 Find: [F^-1(8)]'?

I'm in AP Calculus

Update:

There is no f(x)

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• Anonymous

Are you SURE F'(3) = 7????

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• Presuming F is one-to one, if F(3) = 8, then we take the functional inverse of both sides to obtain

F^(-1)(F(3)) = F^(-1)(8)

3 = F^(-1)(8)

I have absolutely no clue what role the derivatives played. But at first glance, this is my answer.

EDIT: I missed the apostrophe at the end. I'm not sure how to do the rest of the question.

• 1. F^-1(x) is not 1/F(x) It is reverse function or it will use the notation 1/F(X).

2. Use Chain Rule twice.

Use Chain Rule:

u=g(x)

y=f(u)

So y = f(g(x))

Then dy/dx = dy/du * du/dx

[f(g(x)]' = f'(g(x)) * g'(x)

g'(x) = [f(g(x))]' / f'(g(x))

Let's make G(x) be F^-1(x) and we got:

[G(x)]' = [F(G(x))]' / [F(x)]'

-> [F^-1(x)]' = [F(F^-1(x)]'/[F(x)]'

-> [F^-1(8)]' = [F(F^-1(8))]'/[F(8)]'

-> [F^-1(8)]' = [F(3)]'/[F(8)]'

-> [F^-1(8)]' = 7/[F(8)]'

Let's find F'(8)

Let u=G(x) = x^3

y = F(u)

F'(G(x)) = G'(x) * F'(x)

F'(G(x)) = 3 * x^2 * F'(x)

Let x = 2

Then F'(8) = 3 * 2^2 * F'(2) = 3 * 4 * 4 = 48

So [F^-1(8)] = 7/[F(8)]' = 7 / 48

• [F^-1(8)]' = 1/F'(3) = 1/7

• [F^-1(8)]' = - F'(8)/F(8)

F(x) = ax^3+bx^2+cx+d

F'(x)= 3ax^2+2bx+c

solving F(x) we get a=1 b=-6 c=16 d=-15

F(x) = x^3-6x^2+16x-15

F(8) =241

F'(8)=48

• Anonymous

F^-1(8) = 3

This is a number, so, its derivative is 0.

Edited:

The following answer sounds right to me. Harry's hypothesis is interesting, but noone knows if f is a 3rd grad polynomials.

Ana

• Ummm, I'm in AP Calculus also, but... gotta tell you, I have no idea... I don't know if we've learned that yet...

• [F^-1(8)]' = 1/F'(3) = 1/7

If you haven't learned it, here is the proof.

F^-1(y) = F^-1[F(x)] = x

Differentiate both side with respect to x,

F^-1'(y)F'(x) = 1