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# 關於兩題微積分mean value theorem的問題 急

show htat the equation x(3次方) - 15x + c =0 has at most one root

in the interval [-2,2]

show that the wquation x(4次方) +4x +c =0 has at most two real roots

基本上兩題方法應該差不多，我只想到用反證法，可是證不出來，有哪位大大能幫幫忙呢 3Q~~

copestone大大

謝謝你的祥解

可是妳寫意見我沒辦法給你點數耶

### 2 Answers

- myisland8132Lv 71 decade agoFavorite Answer
1 Let f'(x)=3x^2-15=0=>x=√5 or -√5

Since if there have more than one root in the interval [-2,2], then there should be at least one x such that f'(x) = 0 by the corollary of Rolle's theorem. However, both -√5 and √5 do not belongs to [-2.2], which is a contradiction.

2 Let f'(x)=4x^3+4=0=>x=-1.

Since there is only one x such that f'(x) = 0. By the corollary of Rolle's theorem, the equation has at most two real roots on the real axis.

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- CopestoneLv 41 decade ago
那就做第一題給你看。

如果 f(x) = x^3 - 15x + c 在 [-2, 2] 之間有兩個相異根 a, b, 且 a < b.

則 f(a) = f(b) = 0，由均值定理，存在 d ∈ (a, b) ⊆ [-2, 2]

使得 f'(d) = [(b) - f(a)]/(b -a) = 0

但對所有 x ∈ [-2, 2], f'(x) = 3x^2 - 15 < 0，矛盾。//

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