# What are the points of inflection of f(x)=cos^2x  on the interval x in [0,2pi]?

Nov 19, 2017

${x}_{1} = \frac{\pi}{4}$
${x}_{2} = \frac{3 \pi}{4}$
${x}_{3} = \frac{5 \pi}{4}$
${x}_{4} = \frac{7 \pi}{4}$

#### Explanation:

You have to calculate the second derivate of $f \left(x\right)$,
$f ' ' \left(x\right) = - 2 {\cos}^{2} x + 2 {\sin}^{2} x$

and inflection point is found by equaling the second derivate to $0$,
$- 2 {\cos}^{2} x + 2 {\sin}^{2} x = 0$

x=(pi·constn(1))/2-pi/4

and we get the points:
${x}_{1} = \frac{\pi}{4}$
${x}_{2} = \frac{3 \pi}{4}$
${x}_{3} = \frac{5 \pi}{4}$
${x}_{4} = \frac{7 \pi}{4}$