# Find the max and min values

1. Jul 11, 2011

### Glissando

1. The problem statement, all variables and given/known data
Find the maximum and minimum values of the function f(x) = 2sin2x-2sinx on the interval [-pi, pi]]

2. Relevant equations
first derivative

3. The attempt at a solution
f'(x) = 4sinxcosx-2cosx
0 = 2cosx (2sinx-1)
0 = 2sinx-1, sinx = 1/2, x = pi/6, 5pi/6
0 = 2cosx, cosx = 0, x = pi/2

I got that MIN = -0.5 (plugged pi/6 and 5pi/6 into the original function) but I can't seem to find where MAX @4 is.

Thank you!

Thank you!

Last edited: Jul 11, 2011
2. Jul 11, 2011

### micromass

Hi Glissando!

You forgot -pi/2 here, no?

Anyway, that would give you 6 critical values (don't forget to add the boundaries of the interval as critical values!!!!):

$$-\pi,-\frac{\pi}{2},\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6},\pi$$

Now you need to calculate f of all these values and find out which one is the greatest!

3. Jul 11, 2011

### Ray Vickson

Put y = sin(x), to get the function 2*y*(y-1) on the interval -1 <= y <= 1. The min occurs at y = 1/2. You have an "end point" max, where the derivative (with respect to y) need not equal zero.

RGV

Last edited: Jul 11, 2011
4. Jul 11, 2011

### Glissando

Thank you (: Figured it out!