Caluclus

Find the absolute maximum and absolute minimum of f on the interval (-1,2]: f(x)=(-x^3+x^2+3x+1)/(x+1)
A. Maximum: (1, -2); Minimum: (-1, 2)
B. Maximum: (1, -2); Minimum: None
C. Maximum: None; Minimum: None
D. Maximum: None; Minimum: (-1, 2)
E. None of these

I got to my minimum and maximum but when you plug in -1 to get the ordered pair it is undefined and I'm not sure what the answer would be.

asked by Anonymous
  1. f(x)=(-x^3+x^2+3x+1)/(x+1)
    This factors and reduces to
    f(x) = (x+1)(-x^2 + 2x + 1)/(x+1)
    = -x^2 + 2x + 1 , x ≠ -1

    f ' (x) = -2x +2 , for all x's except -1
    -2x + 2 = 0
    -2x = -2
    x = 1 , which lies in your given domain of -1 to 2

    so f(1) = -1 + 2 + 1 = 2
    using the 2nd derivative test:
    f ''(x) = -2, so we have a maximum

    (1,2) is the only maximum point
    looks like choice B

    here is a graph of
    f(x)=(-x^3+x^2+3x+1)/(x+1) from Wolfram
    to verify my calculations.

    http://www.wolframalpha.com/input/?i=plot+f%28x%29%3D%28-x%5E3%2Bx%5E2%2B3x%2B1%29%2F%28x%2B1%29+

    posted by Reiny

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