On the segment [-1; 1], the function y = cos x is convex, concave, or has an inflection point.

The second derivative is responsible for the convexity (concavity) and inflection points, we find it:

y ‘= (cos (x))’ = -sin (x).

y ” = (-sin (x)) ‘= -cos (x).

We equate it to zero:

-cos (x) = 0.

The roots of an equation of the form cos (x) = a are determined by the formula:
x = arccos (a) + – 2 * π * n, where n is a natural number.

x = arccos (0) + – 2 * π * n.

x = π / 2 + – 2 * π * n.

There is no x belonging to the given interval, we find the value of the second derivative at the point x0 = 0.

y ” = -cos (0) = -π / 2 <0 hence the function is convex.

Answer: the function is convex.