How do I get the maximum and minimum from this equation

I wouldn't write the first term in terms of 2x, but would leave it in terms of x. Then try factoring the derivative after setting it to zero, and use the factors to solve the equation.
 
MethMath11 said:
F(x) = 4 cos2 x + 4 cos x +10

What I got so far is
F'(x) = 4 sin 2x - 4 sin x
Click to expand...

The derivative of \(\displaystyle \ cos^2(x) \ = \ 2[cos(x)]*[-sin(x)] \ = \ -sin(2x)\).

So, the derivative of F(x) = -4sin(2x) - 4sin(x)

Or, you can leave it as -4*2[sin(x)]*[cos(x)] - 4sin(x) =

-8sin(x)cos(x) - 4sin(x)
 
I just realized something, why not at Max , cos = 1 and at min, cos = -1, you don't have to search for the f'(x) = 0
 
Cos never eqals 1. Maybe cos(x) = 1 but cos never. In fact cos has no meaning at all unlike cos(x)
Your method will work fine. Good job!
Just note that there will be a slight problem if you had F(x) = 4 cos4 x + 4 cos2 x +10
 
MethMath11 said:
I just realized something, why not at Max , cos = 1 and at min, cos = -1, you don't have to search for the f'(x) = 0
Click to expand...
What are you saying the answer is? There are ways to do this without a derivative, but it sounds like you're trying something too simple.

Please follow through with the derivative method, and then check your other method against that. Then graph the function and see if you missed anything.
 
Jomo said:
Cos never eqals 1. Maybe cos(x) = 1 but cos never. In fact cos has no meaning at all unlike cos(x)
Your method will work fine. Good job!
Just note that there will be a slight problem if you had F(x) = 4 cos4 x + 4 cos2 x +10
Click to expand...
I did not realize that you also wanted the min. Your short cut method will work for the max but there will probably be a problem with the min.
 
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