# Homework Help: Extrema 4 part problem - need help

1. Oct 9, 2006

### bballvas

Extrema Problem - Need help

The function f(x) is defined as f(x)=-2(x+2)(x-1)^2 on the open interval (-3,3).

1. Determine the coordinates of the relative extrema of f(x) in the open interval (-3,3)

2. Let g(x) be defined as g(x)= absolute value of f(x) in the open interval (-3,3). Determine the coordinates of the relative maxima of g(x) in the open interval. Explain your reasoning.

3. For what values of x is g'(x) not defined? Explain your reasoning.

4. Find all values of x for which g(x) is concave down. Explain your reasoning.

Thanks for the help.

2. Oct 9, 2006

A good thing to do first would be looking up the definitions of the terms 'extrema', 'concave', etc.

3. Oct 9, 2006

### arildno

1. What are coordinates, and how do we find the relative extrema of a function?
2. What does the absolute value "do" to a number?
3. For what values are the derivative of the absolute value function not defined?
4. What does it mean to be concave down?

4. Oct 9, 2006

### bballvas

I've already done the first part and the second part. I just need help on parts 3 and 4. How do I show that the g'(x) is undefined at -2? Can I do some sort of limit statement?

5. Oct 9, 2006

### bballvas

By the way, my answer to part 1 is:

relative min: (-1,-8)
relative max: (1,0)

and my answer to part 2 is:

relative max (-1,8)

6. Oct 9, 2006

### arildno

What is the derivative of the absolute value function, and where is it not defined?

7. Oct 9, 2006

### bballvas

I broke up g(x) into -2x^3+6x-4 when x≤-2 and 2x^3-6x+4 when x>-2 so the derivative when x≤-2 is -6x^2+6 and the derivative when x>-2 is 6x^2-6. I'm not sure if that's right though. I know g'(x) is undefined at -2.

8. Oct 9, 2006

### HallsofIvy

It's easy to see that if x< -2, f(x) is positive and so
|f(x)|= f(x)= -2(x+2)(x-1)^2. Neither of the relative extrema are less than -2.

If -2< x, then f(x) is negative (0 at x= 1) and so |f(x)|= 2(x+2)(x-1)^2. The relative extrema of that are at the same x as for f(x) but the y values are reversed sign.

Yes, g' is not defined at x= -2. A derivative is not necessarily continuous but does satisfy the "intermediate value property". In particular, it is sufficient to show that the limits of g(x) as x goes to -2 from above and below are not the same. Do you see why g(x) is differentiable at x= 1? "Concave downward" would be like y= -x2. It must have the property that the second derivative is negative. For what values of x is g"(x) negative?

9. Oct 9, 2006

### HallsofIvy

I just noticed that this had also been posted in "Math- Calculus and Analysis". I'm merging the two threads.

10. Oct 9, 2006

### bballvas

I took the second derivative and got:

g"(x)=-12x when x≤-2
g"(x)=12x when x>-2

because it's concave down when g"(x) is negative you only use the second part and then I get that g(x) is concave down on the open interval (-2,0)