Determine the maximum and minimum values of f(x)=2sinx+sin(2x) on the interval 0,2pi

Question

campbell_st · Answer

well the derivative is easy 

$$f'(x) = 2\cos(x) + 2\cos(2x) $$

set f'(x) and solve for x in the given domain

I'd let $$\cos(2x) =2\cos^2(x) - 1$$

so it means you need to solve 

$$4\cos^2(x) + 2\cos(x) - 2 = 0$$

you have a quadratic equation that can be factorised after dividing each term by 2

$$2\cos^2(x) + \cos(x) -1 = 0$$

campbell_st · Answer

now you should be able to find the stationary points, or critical points. You can test the points to determine which is a max and which is a min by using the 2nd derivative test. 

hope this helps

anonymous · Answer

Thanks for the help! This is the only problem that seemed like a blur to me. How can I split this up to find the critical points?

anonymous · Answer

I factored the expression but I do not know what values to test

zepdrix · Answer

Oh you got it factored? Ok good.

So were you able to determine the critical points for x?
Having trouble coming up with test points?

anonymous · Answer

I have (2cos(x)-1)(cos(x)+1) and then I got cosx=1/2 and cosx=-1 and I am stuck here

zepdrix · Answer

Let's look at the second term a moment.

cos x = -1

Think back to your unit circle, when is this true?
What angle? :O

zepdrix · Answer

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