Critical number, max & minimum value

  • Thread starter catjay
  • Start date
  • #1
3
0
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!

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
 

Answers and Replies

  • #2
294
1
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
3
0
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
294
1
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
3
0
I think I know what I did wrong. Thanks!
 

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