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Math Help - I'm so long on this one...

  1. #1
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    I'm so long on this one...

    If f is one-to-one, twice differentiable fucntion with inverse function g, show that

    g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}

    What is this asking me to do?

    Also deduce that if f is increasing and concave upward, hten its inverse function is concave downwards. (But I need to understand the top part before worrying about this)

    I'm so lost...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    If f is one-to-one, twice differentiable fucntion with inverse function g, show that

    g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}

    What is this asking me to do?

    Also deduce that if f is increasing and concave upward, hten its inverse function is concave downwards. (But I need to understand the top part before worrying about this)

    I'm so lost...
    no, it is asking you to derive the formula given. i'll do the hard part.

    Consider the function f \left( g(x) \right)

    since g(x) = f^{-1}(x), we have that:

    f \left( g(x) \right) = x

    differentiating both sides implicitly with respect to x, we get:

    f' \left( g(x) \right)g'(x) = 1 .....................................note, we used the chain rule on the LHS

    differentiating implicitly again, we obtain:

    f'' \left( g(x) \right)g'(x)g'(x) + f' \left( g(x) \right)g''(x) = 0 ......note, we used the product rule (and chain rule) on the LHS

    solving for g''(x), we get:

    g''(x) = - \frac {f'' \left( g(x) \right) \left[ g'(x) \right]^2}{f' \left( g(x) \right)}

    Now, all you need to do, is prove that \left[ g'(x) \right]^2 = \frac 1{ \left[ f' \left( g(x) \right) \right]^2} and you will have the desired result
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  3. #3
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    Ohh

    Since g'(x) = 1/f'(g(x))

    then square that, ok

    Ok thanks! For the second part, I know that since it inverse, then it reflects on the y=x so if it is like a U shape, then reflecting it will make it concave down but how do I prove that using math on paper?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    Ohh

    Since g'(x) = 1/f'(g(x))

    then square that, ok

    Ok thanks! For the second part, I know that since it inverse, then it reflects on the y=x so if it is like a U shape, then reflecting it will make it concave down but how do I prove that using math on paper?
    Hint:

    If f is concave up, what does that say about f''?

    if g is concave down, what does that say about g''?
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  5. #5
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    f'(x) > 0

    f''(x) > 0

    g''(x) < 0

    I can't really deduce anything from this...what else am i missing?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    f'(x) > 0

    f''(x) > 0
    these are what you are given.


    g''(x) < 0
    this is your conclusion based on the formula you have. (g''(x) = - (positive quantity)/(positive quantity) which is obviously negative) mission complete.
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