# Critical number, max & minimum value

1. Mar 31, 2012

### catjay

Hey guys, I seem to have some troubles solving these problems. I was hoping if you guys can give me some insight. Thank you so much for your help!

1. The problem statement, all variables and given/known data
Find the absolute min and max value of each function

2. Relevant equations
f(x) = sin 2x - 2 sin x on (-pi,pi)

3. 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.

1. The problem statement, all variables and given/known data
Find the critical #

2. Relevant equations

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

3. 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

2. Mar 31, 2012

### Poopsilon

For your first problem, 2 cos 2x - 2 cos x ≠ 2 cos x.

For your second I plugged your function into a derivative calculator and got a different numerator, so you made an algebra error there as well.

3. Mar 31, 2012

### catjay

As for the first one, is there any identity I'm not aware of?

As for the second question, the numerator was supposed to be 6-3x. I tried to conjugate it but was unable to get that number.

:/

4. Mar 31, 2012

### Poopsilon

Well you could use the double angle formula on the cos2x which may simplify it some.. you want to set f'(x) equal to zero and solve for x, so maybe play around with that identity or with it as is and see if you can deduce what values of x satisfy the equation.

As for the second I'm not sure why you need to conjugate, you should apply the product rule for derivatives and then simplify by giving all the terms a common denominator.

5. Apr 1, 2012

### catjay

I think I know what I did wrong. Thanks!