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Math Help - Locating stationary points on a curve

  1. #1
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    Locating stationary points on a curve

    The question:

    Locate and identify the stationary points for y = \frac{x}{1+x^2}

    My attempt:

    Using quotient rule...
    \frac{1 - x^2}{(1 + x^2)^2}

    I find when this equals zero, i.e. 1 - x^2 = 0
    Therefore, x = \pm1

    I substitute this into the original eqn to get 1/2 and -1/2. So the points are (1, 1/2) and (-1, -1/2). So far so good.

    Now I want to know the nature of these points. So I go to find the second derivative...

    \frac{-2x(1+x^2)^2 - (1 - x^2)4x(1 + x^2)}{(1 + x^2)^4}

    Looks a bit messy!

    I substitute 1 into the equation and get a negative number, so I gather that (1, 1/2) is a maximum point. However, when I substitute -1, I get 0. My text states that (-1, -1/2) is a local minimum, so it can't be 0, can it?

    Any assistance would be greatly appreciated!
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  2. #2
    Senior Member Dinkydoe's Avatar
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    I'm not sure why you substituted \pm  1 in the second derivative.

    But substituting \pm 1 in \frac{x}{1+x^2} gives points (1,1/2) and (-1,-1/2)

    We know these points are either local or global maxima/minima.
    We only have to check whether there exist x\in \mathbb{R} such that y(x)>1/2 or y(x)< -1/2

    (if you check the graph of y(x) on wolframalpha we actually see that (1,1/2) and (-1,-1/2) are global maxima/minima. Thus I don't understand why your
    text states (-1,-1/2) is a local minimum. )
    Last edited by Dinkydoe; July 21st 2010 at 04:44 AM.
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  3. #3
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    Quote Originally Posted by Dinkydoe View Post
    I'm not sure why you substituted \pm  1 in the second derivative.
    I was trying to do this:
    Second derivative test - Wikipedia, the free encyclopedia
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  4. #4
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    Quote Originally Posted by Glitch View Post
    I haven't checked any of your calculations but if you're using the second derivative test you should have made it your business to know that the test fails if f'' = 0. In such cases the sign test must be used. Regardless, for your problem it's much more sensible to just use the sign test in the first place.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    I haven't checked any of your calculations but if you're using the second derivative test you should have made it your business to know that the test fails if f'' = 0. In such cases the sign test must be used. Regardless, for your problem it's much more sensible to just use the sign test in the first place.
    Ok, I just did it, and it is indeed a minimum point. Thanks, I should have known to check that way. The second derivative is very messy, after all.
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  6. #6
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    Quote Originally Posted by Glitch View Post
    The question:

    Locate and identify the stationary points for y = \frac{x}{1+x^2}

    My attempt:

    Using quotient rule...
    \frac{1 - x^2}{(1 + x^2)^2}

    I find when this equals zero, i.e. 1 - x^2 = 0
    Therefore, x = \pm1

    I substitute this into the original eqn to get 1/2 and -1/2. So the points are (1, 1/2) and (-1, -1/2). So far so good.

    Now I want to know the nature of these points. So I go to find the second derivative...

    \frac{-2x(1+x^2)^2 - (1 - x^2)4x(1 + x^2)}{(1 + x^2)^4}

    Looks a bit messy!

    I substitute 1 into the equation and get a negative number, so I gather that (1, 1/2) is a maximum point. However, when I substitute -1, I get 0.
    No, you don't! When x= -1, this becomes
    \frac{-2(-1)(1+(-1)^2)^2 - (1 - (-1)^2)4(-1)(1 + (-1)^2)}{(1 + (-1)^2)^4}
    \frac{(2)(4)}{16}= \frac{1}{2}> 0.

    I wonder it you neglected squaring -1 before adding it to 1?

    My text states that (-1, -1/2) is a local minimum, so it can't be 0, can it?

    Any assistance would be greatly appreciated!
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    No, you don't! When x= -1, this becomes
    \frac{-2(-1)(1+(-1)^2)^2 - (1 - (-1)^2)4(-1)(1 + (-1)^2)}{(1 + (-1)^2)^4}
    \frac{(2)(4)}{16}= \frac{1}{2}> 0.

    I wonder it you neglected squaring -1 before adding it to 1?
    Hmm, I made sure I put brackets on my calculator to ensure that it would square properly. Perhaps I made a typo. Thanks.
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  8. #8
    A Plied Mathematician
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    You can always use this mnemonic device for remembering the Second Derivative Test:

    Locating stationary points on a curve-second-derivative-test-mnemonic.jpg

    If the second derivative is negative (-), you're concave down (first face). If the second derivative is positive (+), you're concave up (second face). If the second derivative is zero (0), you could be concave down, concave up, or have an inflection point (third, confused face).

    Cheers!
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