TITLE: Trigonometry inflection point
QUESTION [7 upvotes]: 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

REPLY [10 votes]: 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 $0$.

REPLY [3 votes]: 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 $\delta$.