Finding Relative Extrema and Points of Inflection

**Bashyboy** · May 13th 2012, 05:02 PM

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?

**Prove It** · May 13th 2012, 05:11 PM

Is it asking for STATIONARY points of inflexion?

**skeeter** · May 13th 2012, 05:13 PM

your second derivative is wrong.

try again ...

**earboth** · May 13th 2012, 11:35 PM

Originally Posted by Bashyboy

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 $f''(x)=0$

**Bashyboy** · May 14th 2012, 07:16 AM

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?

**skeeter** · May 14th 2012, 07:22 AM

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

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