# Thread: Finding Relative Extrema and Points of Inflection

1. ## 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?

2. ## Re: Finding Relative Extrema and Points of Inflection

Is it asking for STATIONARY points of inflexion?

3. ## Re: Finding Relative Extrema and Points of Inflection

try again ...

4. ## Re: Finding Relative Extrema and Points of Inflection

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?

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

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

5. ## Re: Finding Relative Extrema and Points of Inflection

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?

6. ## Re: Finding Relative Extrema and Points of Inflection

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