# Thread: How to find absolute extrema.

1. ## How to find absolute extrema.

Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive

But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)

So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well

2. ## Re: How to find absolute extrema.

Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
The title of the thread says it all: absolute extrema.
Is $\displaystyle (1,1)$ an absolute extreme point?

3. ## Re: How to find absolute extrema.

Originally Posted by Chaim
Well I have a problem like:
f(x) = 3x2/3 - 2x, Interval: [-1, 1]
So this is what I did first:
Found the derivative: 2x-1/3-2
0 = 2x-1/3-2
x = 1
OR
f'(x) = 2/x1/3 - 2
x does not equal 0
Critical numbers: x = 1, x = 0
Used a chart:
x [-1, 0] 0 (0, 1) 1
f(x) negative undefined positive 0
Minimum occurs at x=0 since it goes from negative to positive

But I'm a bit confused on the maximum
I plugged in the points f(-1) = 3(-1)2/3 - 2(-3)
=5, so the point is (-1, 5)
Then I plugged in 1
f(1) = 3(1)2/3 - 2(1)
=1, so the point is (1, 1)

So I'm a bit confused, since at the back of my book,
The minimum is: (0, 0)
While the only maximum is (-1, 5)
So how come (1, 1) didn't get included as well
Always keep the original function in mind as you choose critical points (or anything else for that matter.)

So. Critical points.
Solve $\displaystyle 2x^{2/3} - 2x = 0$ for critical points.
Factor the x:
$\displaystyle x \left ( 2x^{-1/3} - 2 \right ) = 0$

The solutions are $\displaystyle x = 0$ or $\displaystyle \left ( 2x^{-1/3} - 2 \right ) = 0$

You can solve this in any way you like, and I also get $\displaystyle x = \pm 1$.

So we have 3 critical points $\displaystyle x = -1,~0~, 1$

We also need to check two more things. We need to see if the derivative has any critical points. In this case that would be anything were the derivative does not exist or is 0. You already have critical points at $\displaystyle x = \pm 1$ so these don't give anything new. The other thing to check is the value of the function (and its derivative) are at the ends of your domain. Again, this doesn't give anything new.

You did the rest of it correctly.

-Dan

4. ## Re: How to find absolute extrema.

The point (1,1) is a local maximum, but it's not the absolurte maximum within teh range [-1,1]. Look at the attached graph - clearly the value of the function at x=-1 is greater than at x=1:

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### Locate the absolute extrema of the function on the closed interval. y = 3x2/3 − 2x, [−1, 1]

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