1. ## Absolute minimum question...

I'm having a ton of trouble trying to find the absolute minimum of this function on the domain [0,2pi]

$\displaystyle cosx(1-2sinx)$

I know I must set this equation equal to 0 and solve for the values where it is 0 but I keep getting stuck. Any help would be greatly appreciated.

2. Wow that one is a pain. Easier to just solve it numerically in your calculator. Take the derivative with respect to x and then plot it as well as the original function. Get the calculator to find the root of the derivative function near the point where you see the original function is minimum.

I think the exact number is given by:

$\displaystyle 2 \pi -\cos ^{-1}\left(-\frac{1}{4} \sqrt{\frac{1}{2} \left(15+\sqrt{33}\right)}\right) \approx 3.77646$

3. I think that did it for me, thanks!

I've got one more that's been bothering me though. These critical points really get to me...

I have to find all the critical points of f when $\displaystyle f(x)=x^{4/5}(x-5)^2$

Any thoughts?

4. Originally Posted by rust1477
I'm having a ton of trouble trying to find the absolute minimum of this function on the domain [0,2pi]

$\displaystyle cosx(1-2sinx)$

I know I must set this equation equal to 0 and solve for the values where it is 0 but I keep getting stuck. Any help would be greatly appreciated.
No, you don't set that equation equal to 0. You set the derivative of that function equal to 0. that turns out to be a much easier equation!

Originally Posted by rust1477
I think that did it for me, thanks!

I've got one more that's been bothering me though. These critical points really get to me...

I have to find all the critical points of f when $\displaystyle f(x)=x^{4/5}(x-5)^2$

Any thoughts?
Well, yes: the critical points of a function are defined as places where the derivative is either 0 or does not exist. What is the derivative of f(x)? (There are 3 critical points.)