Question

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

Answer 1

how do you know?

Answer 2

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

Answer 3

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 = \[x =\pi/4, 3\pi/4\]

Answer 4

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

Answer 5

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