Inflection point confusion

  • #1

Homework Statement


[tex] f(x) = x^{4} - 2x^{2} + 3[/tex]

Find the intervals of concavity and the inflection points.

Homework Equations


[tex]f''(x) = 4(3x^{2}-1)[/tex]



The Attempt at a Solution


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

Thanks
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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22/9 isn't an "inflection point" but then neither are [itex]1/\sqrt{3}[/itex] nor [itex]1/\sqrt{3}[/itex]! They are not points! What your book is saying is that [itex]\left(1/\sqrt{3}, 22/9\right)[/itex] and [itex]\left(-1/\sqrt{3}, 22/9\right)[/itex] are the inflection points. That is, when x is [itex]1/\sqrt{3}[/itex] or [itex]-1/\sqrt{3}[/itex], y is equal to 22/9.
 
  • #3
22/9 isn't an "inflection point" but then neither are [itex]1/\sqrt{3}[/itex] nor [itex]1/\sqrt{3}[/itex]! They are not points! What your book is saying is that [itex]\left(1/\sqrt{3}, 22/9\right)[/itex] and [itex]\left(-1/\sqrt{3}, 22/9\right)[/itex] are the inflection points. That is, when x is [itex]1/\sqrt{3}[/itex] or [itex]-1/\sqrt{3}[/itex], y is equal to 22/9.

Of course! Thanks for your help
 

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