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Math Help - Differentiability analysis

  1. #1
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    Differentiability analysis

    How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

    {\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0?
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  2. #2
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    Quote Originally Posted by FractalMath View Post
    How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

    {\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0?
    You don't. You can't. It's not true. If there were a differentiable function satisfying that equation, then, differentiating both sides, 5f(x)^4f'(x)+ f'(x)= f'(x)(5f(x)^4+ 1)= 0. So either f'(x)= 0, for all x, or 5f(x)^4+ 1= 0. If f'(x)= 0 for all x, then f(x) is a constant, c, such that c^5+ c+ x= 0 or c^5+ c= -x which is impossible: the right side changes as x changes and the left side is a constant. However, all terms in 5f(x)^4+ 1 are non-negative: their sum is 0 only if 5f(x)^4= -1 which again says that f(x) is a constant.
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    Quote Originally Posted by HallsofIvy View Post
    You don't. You can't. It's not true. If there were a differentiable function satisfying that equation, then, differentiating both sides, 5f(x)^4f'(x)+ f'(x)= f'(x)(5f(x)^4+ 1)= 0. So either f'(x)= 0, for all x, or 5f(x)^4+ 1= 0. If f'(x)= 0 for all x, then f(x) is a constant, c, such that c^5+ c+ x= 0 or c^5+ c= -x which is impossible: the right side changes as x changes and the left side is a constant. However, all terms in 5f(x)^4+ 1 are non-negative: their sum is 0 only if 5f(x)^4= -1 which again says that f(x) is a constant.
    What you say isn't true.

    Differentiating both sides we get 5{\left( {f\left( x \right)} \right)^4}f'\left( x \right) + f'\left( x \right) + 1 = 0. You're missing the +1.
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    Quote Originally Posted by FractalMath View Post
    What you say isn't true.

    Differentiating both sides we get 5{\left( {f\left( x \right)} \right)^4}f'\left( x \right) + f'\left( x \right) + 1 = 0. You're missing the +1.
    Ouch! I need to have my eyes checked. Yes, you are completely correct (Actually it was the original "x" that has derivative one that I missed.)
    So, what can you say about that equation?
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