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Thread: finding derivative of an inverse function given f(x) = x^5 +3x +6

  1. #1
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    finding derivative of an inverse function given f(x) = x^5 +3x +6

    given that f(x) = x^5 +3x +6 is one to one, find (f^-1)' (10) (the derivative of the inverse function at x=10)

    I know that the derivative of an inverse function = 1/f'(f^-1 (x))
    f'(x) = 5x^4+3

    However, I am not sure about how to find the value of f^-1 (10)


    i know that point (10, y) on the inverse is the point (y, 10) on f(x)
    So if I can isolate for the x value in the function f(x) = x^5 +3x +6 I will know (10, y) and I will know what f^-1 (10) is

    however, f(x) is a quintic function, and I don't know how to isolate for the x value.

    I've concluded there must be an easier way to solve this that I am not seeing, so any guidance would be much appreciated.

  2. #2
    I'm assuming you would solve for the derivative as normal.

    Then since its f prime to the negative 1 power you would just change it to 1 / f prime. Then solve for your point at x = 10

    If its f to the negative 1 prime.. then invert and use the quotient rule.

  3. #3
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    Quote Originally Posted by fresh View Post
    given that f(x) = x^5 +3x +6 is one to one, find (f^-1)' (10) (the derivative of the inverse function at x=10)

    I know that the derivative of an inverse function = 1/f'(f^-1 (x))
    f'(x) = 5x^4+3

    However, I am not sure about how to find the value of f^-1 (10)


    i know that point (10, y) on the inverse is the point (y, 10) on f(x)
    So if I can isolate for the x value in the function f(x) = x^5 +3x +6 I will know (10, y) and I will know what f^-1 (10) is

    however, f(x) is a quintic function, and I don't know how to isolate for the x value.

    I've concluded there must be an easier way to solve this that I am not seeing, so any guidance would be much appreciated.
    I cannot tell you a general approach to solving quintics, but the integer root theorem lets you solve this one with ease.

    [tex]x^5 + 3x + 6 = 10 \iff x^5 + 3x - 4 = 0.[/tex]

    Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1.

    [tex]1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.[/tex]

    [tex]\therefore f(1) = 10 .[/tex]

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    Translate the graph at (1,10) to (0,0)

    P(x) = x^5+3x+6
    P(x+1)-10 = (x+1)^5+3(x+1)+6-10
    P(x+1)-10 = x^5+5x^4+10x^3+10x^2+8x
    P(x+1)-10 ~ 8x near the origin

    P(x+1)-10 at (0,0) has the same slope as P(x) at (1,10)
    P'(1) = 8
    InvP(10) = -1/8
    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

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    Quote Originally Posted by Bob Brown MSEE View Post
    P(x) = x^5+3x+6
    P(x+1)-10 = (x+1)^5+3(x+1)+6-10
    P(x+1)-10 = x^5+5x^4+10x^3+10x^2+8x
    P(x+1)-10 ~ 8x near the origin

    P(x+1)-10 at (0,0) has the same slope as P(x) at (1,10)
    P'(1) = 8
    InvP(10) = -1/8
    I do not understand this answer.

    What is the relevance of P(x + 1)? There is no difficulty in figuring out the derivative of P(x).

    [tex]P'(x) = 5x^4 + 3 \implies P'(1) = 8 \implies \dfrac{1}{P'(1)} = \dfrac{1}{8}.[/tex]

    The question asked was how to find [tex]P^{-1}(10).[/tex]

    Once you know that [tex]P^{-1}(10) = 1,[/tex] the problem is trivial.

    Clearly I am missing something subtle.

  6. #6
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    Quote Originally Posted by JeffM View Post
    I cannot tell you a general approach to solving quintics, but the integer root theorem lets you solve this one with ease.

    [tex]x^5 + 3x + 6 = 10 \iff x^5 + 3x - 4 = 0.[/tex]

    Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1.

    [tex]1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.[/tex]

    [tex]\therefore f(1) = 10 .[/tex]
    Thank you! the problem is really quite simple with the integer root theorem applied.

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    Post Jeff

    Quote Originally Posted by JeffM View Post

    The question asked was how to find [tex]P^{-1}(10).[/tex]

    Once you know that [tex]P^{-1}(10) = 1,[/tex] the problem is trivial.

    Clearly I am missing something subtle.
    Hi Jeff,

    Its good to talk with you again, I've been away.
    On this post, (my bad) I miss read the OP and had the impression that he had guessed by inspection that (1,10) was on the graph. If so, then ...
    STEPS:
    1) Show that (1, 10) is on the polynomial.
    2) Calculate the slope at (1,10)

    My solution was simply a way to do both steps at once.
    HOWEVER I went back to re-read the post -- he did not have [tex]P^{-1}(10).[/tex] -- Clearly you did not miss something subtle

    Thanks for catching my error.
    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

  8. #8
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    Bob

    Ahh I see. Not a problem. I really did think there might have been something subtle going on.

    Thanks for following up.

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