What are the points of inflection of $f (x) = {cos}^{2} x$ $f(x)=cos^2x$ on the interval $x \in [0, 2 π]$ $x in [0,2pi]$ ?

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Answer 1 · 2017-11-19T11:40:09

$x_{1} = \frac{π}{4}$ $x_1=pi/4$
$x_{2} = \frac{3 π}{4}$ $x_2=(3pi)/4$
$x_{3} = \frac{5 π}{4}$ $x_3=(5pi)/4$
$x_{4} = \frac{7 π}{4}$ $x_4=(7pi)/4$

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

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

What are the points of inflection of f(x)=cos^2x f(x)=cos2xf(x)=cos^2x on the interval x in [0,2pi]x∈[0,2π]x in [0,2pi]?