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Math Help - Question on monotonicity problem

  1. #1
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    Question on monotonicity problem

    So here's my problem:

    Find where the given function is increasing and decreasing.

    f(x)=\frac{x-1}{x^2}

    So f'(x)=\frac{(x-1)(2x)-(x^2)(1)}{(x^2)^2)}

    =\frac{2x^2-2x-x^2}{(x^2)^2)}
    =\frac{x^2-2x}{(x^2)^2}

    So to find the split points would I solve 2x^2-2x>0 and 2x^2-2x<0 ?
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  2. #2
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    Solve f'(x) = 0, this means 2x^2-2x=0 which is a 2nd degree polynomial with a = 2 > 0, this means 2x^2-2x is negative between its roots [x1, x2] and positive for the rest (-inf, x1) U (x2, +inf), if a < 0 the sign would be reversed.

    if f`(x) < 0 then f(x) is decreasing on that interval, f`(x) > 0 increasing on that interval

    i forgot to mention (x^2)^2=x^4 is always positive so it doesn't interfere with the sign of your derivative
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Can solved either with l'Hopital's rule.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    My mistake! Sorry!
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  5. #5
    A Plied Mathematician
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    I think you have a sign error in your derivative. For the quotient rule, I always use this mnemonic:

    "lo dee-hi minus hi dee-lo over the square of what's below". That is,

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

    If you compare this with how you took your derivative, I think you'll see the error. And this error will affect your results, because all your answers will be reversed from what they should be.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by ascendancy523 View Post
    So here's my problem:

    Find where the given function is increasing and decreasing.

    f(x)=\frac{x-1}{x^2}

    So f'(x)=\frac{(x-1)(2x)-(x^2)(1)}{(x^2)^2)}

    ...
    The derivative of a fraction is...

    \displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)\ f'(x) - f(x)\ g'(x)}{g^{2}(x)} (1)

    ... so that is...

    \displaystyle \frac{d}{dx} \frac{x-1}{x^{2}} = \frac{2-x}{x^{3}} (2)

    ... and the function \displaystyle \frac{x-1}{x^{2}} is increasing for 0 < x < 2 ...

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by chisigma View Post
    The derivative of a fraction is...

    \displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)\ f'(x) - f(x)\ g'(x)}{g^{2}(x)} (1)

    ... so that is...

    \displaystyle \frac{d}{dx} \frac{x-1}{x^{2}} = \frac{2-x}{x^{3}} (2)

    ... and the function \displaystyle \frac{x-1}{x^{2}} is increasing for 0 < x < 2 ...

    Kind regards

    \chi \sigma
    Alright, so f(x) is increasing on [0,2] and decreasing on ( -\infty,0] \cup[2, \infty)?

    Thank you everyone for the help!
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  8. #8
    A Plied Mathematician
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    You're very welcome. Have a good one!
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