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Homework Help: Extrema 4 part problem - need help

  1. Oct 9, 2006 #1
    Extrema Problem - Need help

    Please help me with this problem if you can:

    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. jcsd
  3. Oct 9, 2006 #2

    radou

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    A good thing to do first would be looking up the definitions of the terms 'extrema', 'concave', etc.
     
  4. Oct 9, 2006 #3

    arildno

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    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?
     
  5. Oct 9, 2006 #4
    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?
     
  6. Oct 9, 2006 #5
    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)
     
  7. Oct 9, 2006 #6

    arildno

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    What is the derivative of the absolute value function, and where is it not defined?
     
  8. Oct 9, 2006 #7
    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.
     
  9. Oct 9, 2006 #8

    HallsofIvy

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    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?
     
  10. Oct 9, 2006 #9

    HallsofIvy

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    I just noticed that this had also been posted in "Math- Calculus and Analysis". I'm merging the two threads.
     
  11. Oct 9, 2006 #10
    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)
     
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