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Can anyone help me find the points of inflection in the following function in the interval between $0$ and $2\pi$

$f(x)=\sqrt{2}x^2-4\sin(x)$

for my first derivative I got

$f'(x)=2\sqrt{2}x-4\cos(x)$

$f''(x)=2\sqrt{2}+4\sin(x)$

$\sin(x)=\frac{-\sqrt{2}}{2}$ inflection points would be $x=\frac{5\pi}{4},\frac{7\pi}{4}$

but would this be correct

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    $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – George V. Williams Mar 17 '13 at 21:05
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    $\begingroup$ These are the inflection points. In principle, you also need to chow that concavity actually changes at these points. Setting the second derivative equal to $0$ and solving is not enough. For example, $x^4$ has second derivative $0$ at $0$, but concavity doesn't change, $0$ is not an inflection point. $\endgroup$ – André Nicolas Mar 17 '13 at 21:10
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    $\begingroup$ Yes I will make a number line to show how they change. $\endgroup$ – Fernando Martinez Mar 17 '13 at 21:11
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That is correct, but one more remark is worth making:

An inflection point is not merely a point where the second derivative is $0$, but rather is a point where the second derivative changes from positive to negative or vice-versa.

For example, if $g(x)=x^4$, then $g''(x)=0$ when $x=0$, but that's not an inflection point since $g''(x)$ is positive if $x$ is on either side of $0$.

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  • $\begingroup$ I see so on only on one of those points does the line change between concave up and concave down $\endgroup$ – Fernando Martinez Mar 17 '13 at 21:10
  • $\begingroup$ I see that the graph is concave up from negative infinity to 5pi/4 and from 7pi/4 to infinity it is concave down from 5pi/4 to 7pi/4 $\endgroup$ – Fernando Martinez Mar 17 '13 at 21:15
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Hint: $x_0$ is an inflection point if $f'(x_0)=0$ and $f''$ change sign in the intervals $(x_0-\delta,x_0)$ and $(x_0,x_0+\delta)$ for some $\delta$.

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