Why Are Inflection Points Misidentified in Calculus Problems?

[PsychStudent]

Homework Statement

f(x) = x^{4} - 2x^{2} + 3

Find the intervals of concavity and the inflection points.

Homework Equations

f''(x) = 4(3x^{2}-1)

The Attempt at a Solution

f''(x) is zero at \pm\frac{1}{\sqrt{3}}
I've found the correct intervals of concavity, which are (-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty)
I would expect the inflection points to be \pm\frac{1}{\sqrt{3}}, which is partly correct, but the answer the book gives is \pm\frac{1}{\sqrt{3}}, \frac{22}{9}.
I can't see how \frac{22}{9} could be an inflection point. It does not equal zero when plugged into the second derivative.

Thanks

 

[HallsofIvy]

22/9 isn't an "inflection point" but then neither are 1/\sqrt{3} nor 1/\sqrt{3}! They are not points! What your book is saying is that \left(1/\sqrt{3}, 22/9\right) and \left(-1/\sqrt{3}, 22/9\right) are the inflection points. That is, when x is 1/\sqrt{3} or -1/\sqrt{3}, y is equal to 22/9.

 

[PsychStudent]

22/9 isn't an "inflection point" but then neither are 1/\sqrt{3} nor 1/\sqrt{3}! They are not points! What your book is saying is that \left(1/\sqrt{3}, 22/9\right) and \left(-1/\sqrt{3}, 22/9\right) are the inflection points. That is, when x is 1/\sqrt{3} or -1/\sqrt{3}, y is equal to 22/9.

Of course! Thanks for your help