發問時間: 科學數學 · 1 0 年前

微積分問題微積分問題

請盡量寫詳盡、正式一點,記得寫上引用的公式

1.Show that the equation X^3 - 15X + c = 0 has at most one root in the

interval [-2,2 ].

2.If f(1)=10 and f'(x)>2 for 1< x< 4,how small can f(4) possible be?

3.Suppose that f and g are continuous on [a,b] and differentiable on (a,b).

Suppose also that f(a)=g(a) and f'(x)<g'(x) for a<x<b. Prove that

f(b)<g(b).(hint:Apply the Mean Value Theorem to the function h= f-g )

4.Use the Mean Value Theorem to prove the inequality

│sina - sinb│小於等於 │a - b│for all a and b.

2 個解答

評分
  • 1 0 年前
    最佳解答

    您用英文問的,我用英文答。

    1.

    f(X) = X^3 -15X + c,

    f(-2) = -8 + 30 = 22,

    f(2) = 9 - 30 = -22.

    The first order derivative f'(X) = 3X^2 - 15, which is always negative in the interval [-2, 2].

    By the fact that one constantly decreasing function starts from 22 and ends with -22 can only pass through 0 once, f(X) has at most one root in the interval [-2,2] #

    2. By Mean Value Theorem, (f(4) - f(1)) / (4-1) = f'(c), where c is in [1,4].

    Use he fact that f'(c) > 2 , the equation above can be written as:

    (f(4) - f(1)) / 3 > 2 => f(4) - f(1) > 6.

    Given f(1) = 10, f(4) > 6 + f(1) = 16

    Answer: f(4)>16

    3.

    Let h(x) = f(x) - g(x), and h(x) is continuous on [a,b] and differentiable on (a,b).

    h'(x) = (f(x)-g(x))' = f'(x) - g'(x).

    By Mean Value Theorem, there exists a c in [a,b], such that

    h'(c) = f'(c) - g'(c) = {[f(b) - g(b)] - [f(a) - g(a)]} / (b - a).

    We know f(a) = g(a), so f(a) - g(a) = 0.

    Then the equation above becomes:

    [f(b) - g(b)] / (b - a) = f'(c) - g'(c) < 0 ( given f'(c) < g'(c) )

    (b-a)>0 by definition, so that f(b) - g(b) < 0 => f(b) < g(b) #

    4. By Mean Value Theorem,

    │sin(a) - sin(b)│/ │a - b│ =│sin'(c)│=│cos(c)│

    Because | cos(c) | <= 1 for all c by definition,

    │sin(a) - sin(b)│/ │a - b│<= 1,

    and │sin(a) - sin(b)│<= │a - b│#

    參考資料: myself
  • 1 0 年前

    1. 設f(x)=x^3-15x+c

    f'(x)=3x^2-15=3(x^2-5)<0...因x在[-2,2].所以遞減.

    設f(x)在[-2,2]有2相異實根x1.x2

    f(x1)=f(x2)=0,f屬於c[-2,2],f屬於(-2,2).

    則存在C屬於(-2,2)使得f'(C)=0

    但f'(x)=3x^2-15=3(x^2-5)<0...與x屬於(-2,2)矛盾

    ie f(x)在[-2,2]恰有一實根

    2.By MVT

    存在c,使得f'(c)=[f(4)-f(1)]/(4-1)=[f(4)-10]/3>2

    f(4)-10>6

    f(4)>16

    3.設h(x)=f(x)-g(x)

    因為當a<x<b,f'(x)<g'(x).

    所以h'(x)=f'(x)-g'(x)<0.

    By MVT

    存在a<c<b,使得h'(c)={[f(b)-g(b)]-[f(a)-g(a)]}/(b-a)

    因為f(a)=g(a),h'(c)<0.

    所以h'(c)=[f(b)-g(b)]/(b-a)

    f(b)-g(b)<0

    f(b)<g(b).

    4.設f(x)=sinx

    By MVT

    存在c屬於(a,b)

    使得f'(c)=(sinb-sina)/(b-a)=sin'c=cosc

    │cosc│<=1,所以│(sinb-sina)/(b-a)│<=1

    │sinb-sina│<=│b-a│

    │sina-sinb│<=│a-b│.

    MVT=Mean Value Theorem

    希望能幫到你.不懂再問我吧!

    參考資料: 我...數學系
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