Dante 發問時間: 科學數學 · 1 0 年前

微積分 平均值 (10 point!!)

For what value of a , m , and b does the function

------- ┌ 3 x=0

for=> │ -x^2+3x+a 0<x<1

------- └ mx+b 1<=x<=2

satisfy the hypotheses of the Mean Value Theorem on the interval [0,2]

?

1 個解答

評分
  • 1 0 年前
    最佳解答

    If we need the Mean Value Theorem to hold for the interval [0, 2], f(x) should:

    (1) Continuous on the interval [0, 2] (Closed interval), and

    (2) Differentiable on the interval (0, 2) (Open interval)

    Therefore, for (1), we have:

    f(0) = 3

    So, lim (x-> 0+) f(x) = 3

    lim (x-> 0-) (-x^2 + 3x + a) = 3

    a = 3

    Then, for (2), we have:

    f'(x) = -2x + 3 for 0 < x < 1 and

    f'(x) = m for 1 <= x <= 2.

    So for f being differentiable at x = 1, we have:

    Left hand derivative = Right hand derivative

    i.e. -2(1) + 3 = m

    m = 1

    Again, back to condition (1) for x = 1:

    f(1) = m + b = b + 1

    So, lim (x-> 1-) f(x) = b + 1

    lim (x-> 1+) (-x^2 + 3x + 3) = b + 1

    b + 1 = 5

    b = 4

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