Question

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

Answer 1

`x_1=pi/4`
`x_2=(3pi)/4`
`x_3=(5pi)/4`
`x_4=(7pi)/4`

Explanation:

You have to calculate the second derivate of `f(x)`,
`f''(x)=-2cos^2x+2sin^2x`

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

`x=(pi·constn(1))/2-pi/4`

and we get the points:
`x_1=pi/4`
`x_2=(3pi)/4`
`x_3=(5pi)/4`
`x_4=(7pi)/4`