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Thread: 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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    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: $\displaystyle \left( f^{-1} \right)'(x) = \frac 1{f' \left( f^{-1}(x) \right)}$


    here, $\displaystyle h(x) = f^{-1}(x)$
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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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  4. #4
    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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    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 $\displaystyle \left( \frac 1{f(x)} \right)'$ not $\displaystyle \frac 1{f'(x)}$

    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: $\displaystyle h'(x) = \frac 1{f'(h(x))}$

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

    first we find $\displaystyle f'(x)$:

    $\displaystyle f'(x) = 3x^2 + 1$

    ok, that wasn't so hard.

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

    thus we must solve: $\displaystyle x^3 + x = 2$

    it is almost obvious that $\displaystyle x = 1$

    thus, $\displaystyle h(2) = 1$

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

    now we just plug that into our formula:

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