Calc 12

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Let f be the function defined by f(x) = (x^2 + 1)e^-x for -4≤x≤4.

a. For what value of x does f reach its absolute maximum? Justify your answer. b. Find the x-coordinates of all points of inflection of f. Justify your answer.

  • Calc 12 -

    by the product rule ...

    f '(x) = (x^2 + 1)(-1)e^(-x) + 2x e^-x
    = e^-x (-x^2 - 1 + 2x)

    for a max or min, f '(x) = 0
    e^-x (-x^2 - 1 + 2x)
    e^-x = 0 ---> not possible
    or
    x^2 -2x + 1 = 0
    (x-1)^2 =
    x = 1
    f(1) = (1+1)e^-1= 2/e = appr .736

    check the ends since we are given a domain:
    f(-4) = 17*e^4 = appr 928
    f(4) = 17/e^4 = appr .311

    So x = -4 yields the maximum
    while x=1 would yield a local maximum

    points of inflection f ''(x) = 0
    f ''(x) = (x-1)^2 (-e^-x) + 2(x-1)e^-x
    = e^-x(-(x-1)^2 + 2(x-1)
    = e^-x (-x^2 + 4x -3)
    = 0

    x^2 - 4x + 3 = 0
    (x-1)(x-3) = 0

    points of inflection at x = 1 and x = 3

    observe (1, 2/e) is both a turning point and a point of inflection

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