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Math Help - [SOLVED] Stuck on a couple of problems

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Stuck on a couple of problems

    y^3 + y = e^x -x + 9

    a) Show that for every x, there exists only exactly one y.

    b) Let the solution y in (a) be y = f(x). Show that this function is C^1 locally.

    c) With the previous two answers in mind, how does it now follow that f \in C^1(R)

    d) What is the domain and range of f.
    Last edited by Spec; May 8th 2008 at 03:19 PM.
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  2. #2
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    Quote Originally Posted by Spec View Post
    y^3 + y = e^x -x + 9

    a) Show that for every x, there exists only exactly one y.

    [snip]
    For a given value of x, e^x - x + 9 = a > 0. And it's simple to show that h(y) = y^3 + y - a is a one-to one function and that there's exactly one solution to h(y) = 0 for all values of x. Therefore .....
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    Quote Originally Posted by Spec View Post
    y^3 + y = e^x -x + 9

    a) Show that for every x, there exists only exactly one y.

    b) Let the solution y in (a) be y = f(x). Show that this function is C^1 locally.

    c) With the previous two answers in mind, how does it now follow that f \in C^1(R)

    d) What is the domain and range of f.
    Let g(x) = e^x -x + 9 and h(y) = y^3 + y. Then g and h are both infinitely differentiable functions. Also, h is strictly increasing, with domain and range both equal to R, so by the inverse function theorem h has a continuously differentiable inverse h^{-1}. The range of g is the half line [10,∞), with the minimum value 10 occurring at x=0.

    If h(y) = g(x) then y=h^{-1}(g(x)), which by the chain rule is a continuously differentiable function of x. The domain of this function is the whole of R, and its range is the half line [h^{-1}(10),\infty).

    [I can't help wondering whether the person who set this question was thinking that g(0)=9 rather than 10. In that case, since h(2)=9, the answer to the last part of the question would come out much more cleanly: the range of f would then be the half line [2,∞).]
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