- #1

- 3

- 0

## Homework Statement

Find the absolute min and max value of each function

## Homework Equations

f(x) = sin 2x - 2 sin x on (-pi,pi)

## The Attempt at a Solution

finding the critical #

f'(x) = 2 cos 2x - 2 cos x

= 2 cos x

set the equation = 0

0 = 2 cos x

x = +pi/2, -pi/2

However, this is not the right answer because when I graphed it, the absolute minimum value was 2pi/3 while the absolute max was squaroot 3/2. I was wondering which step I did wrong? Thank!

I also have another question, in which I'm not sure why I got the answer wrong. The critical # supposed to be x=2,3 while I got x=3,6.

## Homework Statement

Find the critical #

## Homework Equations

f(x) = x (3-x)^(1/2)

## The Attempt at a Solution

f'(x)=[x/(2 * squareroot (3-x)] + [squareroot (3-x)]

finding common denominator:

f'(x) = [x + 2(3-x)]/[2 * squareroot (3-x)]

=(6-x)/[2 * squareroot (3-x)]

Therefore, the critical #s are:

6-x=0 -----> x=6

2 * squareroot (3-x) = 0 ------> x=3