Question

What is the point of inflection for the function f(x)=2cos4x on the interval (0, pi/2)

Answer 1

to find inflection points take the 2nd derivative and make it equal to 0

dy/dx = -8sin4x

d2y/dx2 = -32cos4x

make d2y/dx2 = 0

0 = -32cos4x

0= cos4x

in the unit circle, cos equal 0 at pi/2 and 3pi/2 in that interval, but we have cos4x instead
so:

pi/2 = 4x

x = pi/8

3pi/2 = 4x

3pi/8 = x

answer: pi/8, 3pi/8

Answer 2

x ∈ {π/8, 3π/8}

Explanation:

There are two of them. They lie at the zero-crossings of the function, where ...

4x = π ± π/2

x = π/4 ± π/8 = π/8 or 3π/8