Finding Relative Extrema and Points of Inflection

The function is $\displaystyle f(x) = \frac{x^2}{x^2 - 1}$ The answer key claims there to be no points of inflection, yet I found from my second derivative

$\displaystyle f''(x) = \frac{-10x^2 + 2}{(x^2 - 1)^3}$ I find the points $\displaystyle x = \pm\sqrt{\frac{1}{5}}$ Could it be possible that the answer key is wrong?

Re: Finding Relative Extrema and Points of Inflection

Is it asking for STATIONARY points of inflexion?

Re: Finding Relative Extrema and Points of Inflection

your second derivative is wrong.

try again ...

Re: Finding Relative Extrema and Points of Inflection

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Originally Posted by

**Bashyboy** The function is $\displaystyle f(x) = \frac{x^2}{x^2 - 1}$ The answer key claims there to be no points of inflection, yet I found from my second derivative

$\displaystyle f''(x) = \frac{-10x^2 + 2}{(x^2 - 1)^3}$ I find the points $\displaystyle x = \pm\sqrt{\frac{1}{5}}$ Could it be possible that the answer key is wrong?

Your 2nd derivative is wrong:

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

Obviously there isn't a real solution to $\displaystyle f''(x)=0$

Re: Finding Relative Extrema and Points of Inflection

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Originally Posted by

**Prove It** 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?

Re: Finding Relative Extrema and Points of Inflection

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Originally Posted by

**Bashyboy** 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 $\displaystyle y = x^3$ ... it has a stationary point of inflection at x = 0