[高微] 問兩題題目...

[AM101 (新手)]

１ Prove that the equation x^3 + bx + c = 0 where b > 0 has

　 exactly one solution x in R.





２ Let f:[0,∞)→R be continuous and let f be differentiable on (0,∞).

   Assume f(0)=0 and f(x)→0 as x→∞. Show that there is a

   c in (0,∞) such that f'(c)=0 .



有請高手幫忙解題～～～

感激不盡～～

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推 zombiea     :1.微分恆正, 2. 假設f不恆為零, 另f(y) =k>0 or <0
→ zombiea     :假設大於零, 又f(0)=0 所以存在y_1 使得f(y_1) =k/2
→ zombiea     :y_1 in (0,y), 又f(x)→0 as x→∞ 則存在y_2
→ zombiea     :y_2 in (y,∞) 使得f(y_2) =k/2, Rolle's lemma
→ AM101       :感謝你啦~~ 剛剛發現用 Mean Value Theorem也可以
推 math1209    :對於 2, 我們也可以假設 f'(x) 不為 0. 則根據中值
→ math1209    :定理, f'(x) 只有兩種, 恆正或恆負.
→ math1209    :很明顯地, f'(x) 恆正表示嚴格遞增. 因此不可能會
→ math1209    :產生 f(x)→0 as x→∞. 同理, f'(x) < 0 表示 f(x)
→ math1209    :嚴格遞減...XD
推 azter       :樓上你是使用到Darboux's Theorem?
推 yclinpa     :在[0,∞)上, f 必發生最大值或最小值,在那裏微分為0
→ math1209    :回azter, yes.