Question

y=sinx-(1/2)sin2x for -2pi to 2pi

find critical points and local max and minima

Answer 1

Both are continuous functions, so the sum is continuous.
Let f(x)=sinx-(1/2)sin2x
f'(x)=cos(x)-cos(2x) => f'(x) is continuous and all points of f'(x) defined on the given interval.
Solve for all x on the interval where f'(x)=0 to find the local extrema (and critical points). There are 7 solutions for f'(x)=0, including the two end-points, hence the same number of max-min.

Answer 2

how do i find the asymptotes for this one

Answer 3

There is none, both f(x) and f'(x) are continuous on the interval (in fact, from -inf to +inf).

Answer 4

i only see 2 local max and 2 min in the interval am i missing somting?

Answer 5

Are they symmetrical about x=0?

Answer 6

http://www.wolframalpha.com/input/?i=y%3Dsinx-%281%2F2%29sin2x++

Answer 7

maxima @ 2pi/3 and -4pi/3 right?

Answer 8

minima @
4pi/3 and -2pi/3

Answer 9

do you count -2pi, 0 , and 2pi?

Answer 10

How many points on the graph do you find f'(x)=0?

Answer 11

Yes, those count as well, so there's your answer!

Answer 12

However, check f"(x). If f"(x)=0, then it is NOT a max-min, just a critical point. There is more than one point of this kind.