Respuesta :
Answer:
x ∈ {π/8, 3π/8}
Step-by-step 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
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
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