2cosx+sin2x?

Question

I found that the first derivative is

-4cos(2x)-2sin(x)

Can someone help me find the critical numbers of the derivative? I have all the asymptotes but need to show work.

In my calculator I found there should be critical numbers at pi, 3pi/2, and 3.01626.

I am totally clueless on how to find these. Any help and instructions would be very helpful.

Thank You

Answer 1

f(x) = 2cosx + sin2x
f'(x) = -2sinx + 2cos2x

We have different values for our first derivatives. Nevertheless I'll solve for the critical values.

0 = -2sinx + 2cos2x
0 = -2sinx + 2(1 - 2sin^2(x))
0 = -2sinx + 2 - 4sin^2(x)

Which is a quadratic that I'll rewrite.

4sin^2(x) + 2sinx - 2 = 0
2sin^2(x) + sinx - 1 = 0
(2sinx - 1) (sinx + 1) = 0

Therefore

2sinx - 1 = 0
sinx = -1

2sinx = 1
sinx = -1

sinx = 1/2
sinx = -1

Therefore, presuming restriction of 0 < x <= 2pi,
x = pi/6, 5pi/6
x = 3pi/2

All our values are x = pi/6, 5pi/6, 3pi/2

Answer 2

Um 2cosx does no longer equivalent sin2x(sinx) no remember how puzzling your parentheses are... 2cos0 = 2 sin2*0 = 0 sin0 = 0 2 does no longer equivalent 0 edit: oooh, you had to resolve for x, sorry

Answer 3

mono It's your news?
http://www.osoq.com/funstuff/extra/extra04.asp?strName=mono