how many points of inflection does the graph y=cosx^2 have on the interval (0, pi)

Question

anonymous · Answer

how do you know?

experimentx · Answer

double differentiate => 
y = (cos(x))^2
y' = - 2cos(x)sin(x) = - sin(2x)
y'' = - 2 cos(2x)

for point of inflection, y''=0
or cos(2x) = 0

x=pi/4, 3pi/4

campbell_st · Answer

its 2

f(x) = cos^2x      f'(x) = -2sin(x)cos(x)         f"(x) = -2cos(2x) 

solving -2cos(2x) = 0     

$$2x = \cos^{-1}(0) $$

$$2x = \pi$$

$$x = \pi/2$$

1st and 2nd quadrant

x = $$x =\pi/4, 3\pi/4$$

anonymous · Answer

isn't it just one because it's on the interval of 0 to pi
and 3pi/4 lies in negative x

campbell_st · Answer

3pi/4 < pi.. so it needs to be included