Can someone pleasee help with this problem....

Consider the function f(x)=3cosx−cos(x)^3 for 0

Question

Can someone pleasee help with this problem....

Consider the function f(x)=3cosx−cos(x)^3  for 0<x<2π.

a.) The x-intercepts are .

b.) f′(x)=   -3sin^3(x)

c.) f(x) is increasing on the interval(s) .

d.) f(x) is decreasing on the interval(s) .

e.) f(x) has a local maximum at x = .

f.) f(x) has a local minimum at x = .

g.) f′′(x)=  -9sin^2(x)cosx

h.) f(x) is concave up on the interval(s) .

i.) f(x) is concave down on the interval(s) .

j.) The x-coordinate of the points of inflection are

mertsj · Answer

factor out cosx and get cos(x)[3-cos^2(x)]=y
the x intercepts occur when y is 0 so
cos(x)[3-cos^2(x)]=0
That means cos(x) = 0 or 3-cos^2(x)=0
If cos(x) = 0 x=pi/2 or 3pi/2   if 3-cos^2(x) = 0, cos(x) = +-sqrt3  which is impossible because cos(x) can never be greater than 1 or less than -1.

mertsj · Answer

Is it [cos(x)]^3 or cos(x^3)?????

anonymous · Answer

it's cos(x)^3 it looks like cos^3 x

mertsj · Answer

A.  [cos(x)]^3   OR

B.  cos(x^3)

mertsj · Answer

In other words, is it the cos that is cubed or the angle that is cubed?

anonymous · Answer

yeah cos is cubed