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Math Help - Finding Relative Extrema and Points of Inflection

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    Finding Relative Extrema and Points of Inflection

    The function is f(x) = \frac{x^2}{x^2 - 1} The answer key claims there to be no points of inflection, yet I found from my second derivative
    f''(x) = \frac{-10x^2 + 2}{(x^2 - 1)^3} I find the points x = \pm\sqrt{\frac{1}{5}} Could it be possible that the answer key is wrong?
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    Re: Finding Relative Extrema and Points of Inflection

    Is it asking for STATIONARY points of inflexion?
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    Re: Finding Relative Extrema and Points of Inflection

    your second derivative is wrong.

    try again ...
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    Re: Finding Relative Extrema and Points of Inflection

    Quote Originally Posted by Bashyboy View Post
    The function is f(x) = \frac{x^2}{x^2 - 1} The answer key claims there to be no points of inflection, yet I found from my second derivative
    f''(x) = \frac{-10x^2 + 2}{(x^2 - 1)^3} I find the points x = \pm\sqrt{\frac{1}{5}} Could it be possible that the answer key is wrong?
    Your 2nd derivative is wrong:

    f''(x)=\frac{6x^2+2}{(x^2-1)^3}

    Obviously there isn't a real solution to f''(x)=0
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    Re: Finding Relative Extrema and Points of Inflection

    Quote Originally Posted by Prove It View Post
    Is it asking for STATIONARY points of inflexion?
    I never heard of these stationary points of inflection. What is the difference between points of inflection and stationary points of inflection?
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    Re: Finding Relative Extrema and Points of Inflection

    Quote Originally Posted by Bashyboy View Post
    I never heard of these stationary points of inflection. What is the difference between points of inflection and stationary points of inflection?
    a stationary point of inflection is a point on the curve where f'(x) = 0 , and f''(x) = 0 (and also changes sign)

    look at the graph of y = x^3 ... it has a stationary point of inflection at x = 0
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