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Math Help - inverse function question

  1. #1
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    inverse function question

    the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=

    the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems.

    Thanks for any help I can get
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jst706 View Post
    the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=

    the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems.

    Thanks for any help I can get
    use the formula: \left( f^{-1} \right)'(x) = \frac 1{f' \left( f^{-1}(x) \right)}


    here, h(x) = f^{-1}(x)
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  3. #3
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    every time I try that I get a negative answer, I must be doing something wrong
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jst706 View Post
    every time I try that I get a negative answer, I must be doing something wrong
    do you mind showing your work, so that we can pin-point the areas where you are making mistakes?
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  5. #5
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    Maybe I don't understand the formula...this is what I have been doing.

    1/(x^3+x)...I start with that, then I take the derivative of that

    x^-3+x^-1
    -3x^-4-x^-2

    = 1/ (-3x^4-x^2)

    h'(2)=1/(-3(16)-4)
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jst706 View Post
    Maybe I don't understand the formula...
    it seems you are correct

    this is what I have been doing.

    1/(x^3+x)...I start with that, then I take the derivative of that
    no, that is NOT what you should do. that is finding \left( \frac 1{f(x)} \right)' not \frac 1{f'(x)}

    x^-3+x^-1
    -3x^-4-x^-2

    = 1/ (-3x^4-x^2)
    even if you were correct to do this, this is still wrong. this is not how you find the derivative of a fraction



    ok, let's take this slow.

    the formula is: h'(x) = \frac 1{f'(h(x))}

    we want h'(2), it means we must find h(2) and plug it into f'(x) (this, of course, means we also have to find f'(x))to get f'(h(2)) and then plug it into the formula.

    first we find f'(x):

    f'(x) = 3x^2 + 1

    ok, that wasn't so hard.

    now let's find h(2). since h(x) = f^{-1}(x), this amounts to finding an x such that f(x) = 2

    thus we must solve: x^3 + x = 2

    it is almost obvious that x = 1

    thus, h(2) = 1

    this means that f'(h(2)) = f'(1) = 4.

    now we just plug that into our formula:

    h'(2) = \frac 1{f'(h(2))} = \frac 14
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