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Homework Help: Absolute Min/Max

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the absolute max/min on [-1,3] for y = x^2 * e^-x


    2. Relevant equations



    3. The attempt at a solution
    F'(y) = (2x)(e^-x) + (x^2)(e^-x)
    = xe^-x(2 + x)
    x = 0, -2 (Critical Points)
    F(-2) = 29.56 MAX (CP)
    F(-1) = 10.87
    F(0) = 4
    F(3) = 0.199 MIN (Right Endpoint)

    Can someone please confirm if I am doing this correctly?
     
  2. jcsd
  3. Oct 22, 2007 #2

    Dick

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    The derivative of e^(-x) is -e^(-x), isn't it? Chain rule.
     
  4. Oct 22, 2007 #3
    Ohhhh yeaaaa...thats right. But now how do I reduce it so I can get my critical points?
     
  5. Oct 22, 2007 #4
    F'(y) = (2x)(e^-x) + (x^2)(-e^-x)
    = xe^-x(-2 + x)
    Critical Points = x = 0, 2
     
  6. Oct 22, 2007 #5

    Dick

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    Ok, so far. Continue.
     
  7. Oct 23, 2007 #6
    F'(y) = (2x)(e^-x) + (x^2)(e^-x)
    = xe^-x(2 + x)
    x = 0, -2 (Critical Points)

    Plug CP's and endpoints into original F(x):

    F(-2) = 11.39 MAX (CP)
    F(-1) = 3.72 (ENDPOINT)
    F(0) = 1 MIN (CP)
    F(3) = 9.05 (ENDPOINT)
     
  8. Oct 23, 2007 #7

    Dick

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    Why did -2 turn back into 2 again??
     
  9. Oct 23, 2007 #8
    oops my mistake
     
  10. Oct 23, 2007 #9

    Dick

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    Not the x=-2 critical point, the 2 in your factorization. Why are you making so many mistakes??? The critical points as you said once upon a time are x=0 and x=2. And how can f(0) be 1?? Get a cup of coffee and a clean sheet of paper and start this all over again. Don't cut and paste from previous posts.
     
  11. Oct 24, 2007 #10
    F'(y) = (2x)(e^-x) + (x^2)(-e^-x)
    = e^-x(2x - x^2)
    = xe^-x(2 - x)
    x = 0, 2 (Critical Points)

    I Plugged CP's and endpoints into original F(x):

    F(2) = 0.54 (CP)
    F(-1) = 2.72 MAX (ENDPOINT)
    F(0) = 0 MIN (CP)
    F(3) = 0.45 (ENDPOINT)

    Max = -1
    Min = 0
     
  12. Oct 24, 2007 #11

    HallsofIvy

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    The maximum value is at x= -1. The maximum value itself is e, not "2.7". (No reason to do the extra work-of using a calculator- to get an approximation when you already have an exact value.) The minimum value is at x= 0 and the minimum value is 0.
     
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