# How do you find all points of inflection given y=-2sinx?

Jun 21, 2017

There is a point of inflection whenever $- 2 \sin x = 0$

#### Explanation:

Points of inflection occur when the curve changes concavity. Since this is a sine wave, there are an infinite number of points of inflection.

A function is concave up when the second derivative ($f ' '$) is greater than 0, and concave down when the second derivative is below 0. Critical points, therefore, are when the second derivative equals 0.

Differentiate $y = - 2 \sin x$ to get $y ' = - 2 \cos x$. Differentiate again to get $y ' ' = 2 \sin x$, the original function.

Whenever $- 2 \sin x = 0$, there is a point of inflection. This can be intuitively verified by graphing $y = - 2 \sin x$.

graph{-2sinx [-pi, pi, -3, 3]}

Whenever the curve crosses the x-axis (that is, whenever y=0), the concavity changes.