If then
That's all I can do for you since I don't know whether you mean or .
Either way, there will be two answers depending on the signs of and
John
My calculator said it, I believe it, that settles it
Find the value(s) of x for the following. (Enter your answers as a comma-separated list.)
f(x) = 6 sin^2 x − 6 cos x,
and the interval is 0 is less than or equal to x and x is less than or equal to pi.
(a) f(x) has a local maximum or local minimum.
local maximum: ??
local minimum : ??
(b) f(x)has a global maximum or global minimum.
global max: ??
global min: ??
I do not know how to solve this problem. I understand that I'm supposed to find the derivative first but I don't even know how to take the derivative of sin^2x. Please help me solve this problem!
4 answers
Sorry I think parts of the answer you gave are not visible on the page.
for the derivative, remember the chain rule. In this case, u=sinx and f(u) = u^2, so f'(x) = 2u u', as below.
f = 6 sin^2 x − 6 cos x
f' = 12sinx cosx + 6sinx
(a) max/min where f'=0
6sinx(2cosx+1) = 0
f" = 6cosx + 6cos(2x)
max if f" < 0
min if f" > 0
(b) check extrema for greatest values. Since f is periodic, there won't be many to check.
f = 6 sin^2 x − 6 cos x
f' = 12sinx cosx + 6sinx
(a) max/min where f'=0
6sinx(2cosx+1) = 0
f" = 6cosx + 6cos(2x)
max if f" < 0
min if f" > 0
(b) check extrema for greatest values. Since f is periodic, there won't be many to check.
So the critical points are 0, pi, and 2pi/3 but I don't know where to go from there..?
1 answer