Dom asked in Science & MathematicsMathematics · 1 decade ago

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)

10 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

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

  • liford
    Lv 4
    4 years ago

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  • Puggy
    Lv 7
    1 decade ago

    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 decade ago

    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

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  • [F^-1(8)]' = 1/F'(3) = 1/7

  • 1 decade ago

    [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

    the answer is - 48/241

  • Anonymous
    1 decade ago

    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

  • 1 decade ago

    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...

  • 1 decade ago

    [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

    which leads to

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

  • 1 decade ago

    What is F(x)?

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