# Homework Help: Absolute extrema problem

1. Mar 2, 2005

### jzq

The function is:

$$g(x)=x^2+2x^\frac{2}{3}$$ on [-2,2]

So far I got the derivative as:

$$g'(x)=2x+\frac{4}{3}x^{-\frac{1}{3}}$$

Now, I am stuck at finding the critical #s. I need help.

Last edited: Mar 3, 2005
2. Mar 2, 2005

### Davorak

Wont that be complex when x is negative? How are you going to maximize that?

Last edited: Mar 2, 2005
3. Mar 2, 2005

### jzq

Is there an easier method?

4. Mar 2, 2005

### Davorak

I mean wont the answer have an imaginary part. What criteria are you using to maximize the imaginary part of the answer?

Example:
$${\left( -2 \right) }^{\frac{2}{3}} = -0.793 + 1.37473\ I$$

5. Mar 2, 2005

### jzq

Imaginary? Ooh...I don't think it's suppose to get that complicated. At least not yet. Call me slow but we definately haven't learn that yet. The way I was taught to find critical numbers was set the derivative equal to zero. The cube root makes it difficult for me. How would you simplify the derivative. Once I know that I'll probably be able to take it from there.

I'm not sure what you mean when you say criteria. If the interval is what you mean then here: [-2,2]

Last edited: Mar 2, 2005
6. Mar 2, 2005

### Davorak

Well:
$$g'(x) = 2 x + \frac{4}{3}x^{-\frac{1}{3}} = 0$$
Take $\frac{4}{3}x^{-\frac{1}{3}}$ to the other side of the equation.
use:
$$x^{-1}=\frac{1}{x}$$
Solve for x

Edit:
Does this make sense?

Last edited: Mar 2, 2005
7. Mar 2, 2005

### jzq

So basically you mean: (correct me if I'm wrong)

$$2x=-\frac{4}{3x^\frac{1}{3}}$$

Then divide that by 2? Ok, obviously I'm not very bright in math. Then how would you move the x from $$3x^\frac{1}{3}$$ over?

8. Mar 2, 2005

### Justin Lazear

$$2x=-\frac{4}{3x^\frac{1}{3}}$$

Now multiply both sides by $x^\frac{1}{3}$...

Also, since you're looking for global extrema on a closed interval, you need to test the endpoints.

--J

Last edited: Mar 2, 2005
9. Mar 2, 2005

### Davorak

Yes that is right.
I am being stupid and time to go to bed.

Last edited: Mar 2, 2005
10. Mar 2, 2005

### xanthym

The function g(x) has the following extrema on the Interval [-2, 2]:
Absolute and Relative Minimum at x=(0) where g(0)=(0)
Absolute Maximum at x=(-2) where g(-2)=(7.17)
Absolute Maximum at x=(+2) where g(+2)=(7.17)

Note: g'(x) has NO real values on [-2, 2] where g'(x)=(0). All maxima and minima must be determined "graphically" by computing values of g(x) on [-2, 2].

~~

Last edited: Mar 2, 2005
11. Mar 2, 2005

### jzq

Thanks Xanthym and everyone else!

12. Mar 2, 2005

### Davorak

How are you geting this to be real?

Clarify:

$$(-2)^{-\frac{2}{3}} \in C$$
Where C is the field of Complex numbers.

Last edited: Mar 2, 2005
13. Mar 2, 2005

### xanthym

$$\ \ \ \ g(x) = x^{2} + 2x^{2/3}$$

$$\ \ \ \ g(-2) = (-2)^{2} + 2((-2)^{2})^{1/3} = (7.17)$$

~~

14. Mar 2, 2005

### Davorak

Why is that valid over:
$$g(-2) = (-2)^2 +2((-2)^\frac{1}{3})^2$$

Not that you are claiming such. But if you perform the reverse operation:
$$(((-2)^2)^\frac{1}{3})^\frac{3}{2}=2 \ne -2$$
I thought the root was not considered valid when this was true.

edit:
never mind the ^1/2 takes care of both

15. Mar 2, 2005

### Justin Lazear

An appropriate argument to determine the extrema is to observe the properties of the derivative g'(x).

$$g'(x) = 2x + \frac{4}{3x^{1/3}}$$

Note that the derivative is not defined at x = 0, so therefore it is important to test the point x = 0 as well as the endpoints of the interval, x = ±2.

Also, for $-2 \leq x < 0$, g'(x) < 0. For $0 < x \leq 2$, g'(x) > 0. This implies that local maxima will be found at the endpoints, and the absolute maximum will be one or both of them. It is sufficient to show that g(x) is even, i.e. g(x) = g(-x) to determine that the absolute maximum will be attained at both endpoints. Additionally, the monotonicity of the derivative in the two regions as well as the continuity of g(x) implies that the absolute minimum will be found at x = 0.

There is nothing that needs to be done graphically in this problem.

--J

16. Mar 2, 2005

### dextercioby

There's no imaginary part here.All complex #-s involved are real.What are you dreaming about...?

Daniel.

17. Mar 2, 2005

### Davorak

There are complex numbers which satisfy (-2)^(-2/3). I was just tired and was ignoring the real answers.
I blame my lack of sleep myself.