Identifying Extrema of a Trigonometric Function on an Interval

[Glissando]

Homework Statement

Find the maximum and minimum values of the function f(x) = 2sin²x-2sinx on the interval [-pi, pi]]

Homework Equations

first derivative

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!

 

[micromass]

Hi Glissando!

Homework Statement

Find the maximum and minimum values of the function f(x) = 2sin²x-2sinx on the interval [-pi, pi]]

Homework Equations

first derivative

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

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!

 

[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

 

[Glissando]

Thank you (: Figured it out!