微積分基本定理和roll定理~

1.Let f(x)and f\'(x) be continuous and differentiable functions.

Suppose the curve y=f(x) passes through (0,0),(0,1).Find integral (0~1) f\'(x)dx=?

答=1,請問題目有出錯嗎?

2.Prove that :x^3-3x^2+a=0 has at most one real solution in[0,1] no matter what value

a is.

3 個解答

評分
  • Eric
    Lv 6
    1 0 年前
    最佳解答

    (1)我想應該是:

    Let f and f' be continuous and differentiable functions.

    Suppose the curve y = f(x) passes through the points (0,0) and (1, 1).

    Find ∫(0,1) f'(x) dx.

    We know that f(0) = 0, f(1) = 1.

    ∫(0,1) f'(x) dx = f(1) - f(0) (by the fundamental theorem of calculus)

    = 1 - 0 = 1

    (2)

    Prove that x^3-3x^2+a=0 has at most one real solution in [0,1] regardless of the value of a.

    Proof: Let a be any real number. Let f(x) = x^3-3x^2+a, which has derivative

    f'(x) = 3x^2-6x = 3x(x-2).

    Clearly, f'(x) = 0 only at two points: x = 0 and x = 2.

    Hence f'(x) ≠ 0 on the open interval (0, 1). (*).

    Suppose there exist two solutions u, v with 0 ≤ u < v ≤ 1.

    Then f(u) = f(v) = 0, and thus by Rolle's theorem, there exists a number c, u < c < v, such that f'(c) = 0 -- contradiction.

    Hence, f(x) = 0 has at most one real solution in [0,1].

    ::

  • 1 0 年前

    我誤會了,不論a為何值,至多一個根在[0,1]之間

  • 第二題:他說不管a是什麼…是什麼意思?

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