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Homework Help: Calculus (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

  1. Jun 11, 2010 #1
    "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    1. The problem statement, all variables and given/known data
    Locate the abs extrema on the interval of the function:
    y=t-|t-3| for interval [-1,5]

    2. Relevant equations
    |x|=[tex]\sqrt{x^{2}}[/tex]


    3. The attempt at a solution
    I thought this would essentially be a subtraction rule and chain rule...

    y'=1-((1/(2|t-3|))*2(t-3)*1)
    y'=1-((t-3)/(|t-3|))
    y'=(|t-3|-t+3)/|t-3|

    Critical # at y=3

    t(-1)=-5
    t(3)=3
    t(5)=3

    abs maxima at (5,3) and (3,3)
    abs minimum at (-1,-5)

    Unfortunately, the answer key lists abs max (3,3) and abs min (-1,-1). I don't even get the (-1,-1) since t(-1) is -5...
    If anyone has any guidance, please feel free to let it flow!
     
  2. jcsd
  3. Jun 11, 2010 #2

    Mark44

    Staff: Mentor

    Re: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    It's much easier not to differentiate for this problem. On [-1, 3] the graph is a straight line with slope 2 and y-intercept -3. On [3, 5], the graph is horizontal.

    I agree with you that the answer key is wrong. y(-1) = -5, not -1.
     
  4. Jun 11, 2010 #3
    Re: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    typo? perhaps they want (1,-1) which is indeed a smaller abs(1-abs(1-3))=1 than abs(-1-abs(-1-3))=5 (edit: let me plot this to see...)

    edit: okay after plotting we have a line with positive slope 2, and a horizontal line intesecting at (3,3).

    so the critical points must be: (-1,-5), (3,3), (5,3) but in absolute value the first is the max and the second two are the min...sigh...
     
    Last edited: Jun 11, 2010
  5. Jun 11, 2010 #4
    Re: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    @Mark44:
    I agree that inspection of the graph is easier for this problem, but my professor likes for us to show critical numbers.

    @xaos:
    Once you pointed out the evaluation at 1 (rather than -1), a typo seems to be the most logical conclusion.


    Respectfully to all,

    Aaron
     
  6. Jun 12, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    Re: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    Then show critical numbers- inspection of the graph shows that the derivative is 0 for all x greater than 3 and that the derivative does not exist at x= 3.

     
  7. Jun 12, 2010 #6
    Re: "Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.

    I appreciate the input. I have set about working out analytical that for all real t>3 y'=0.


    Respectfully,

    Aaron
     
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