推 iddee:推 !! 這題很像是政大和中央的某題高微考古題 61.219.178.216 07/13
→ iddee:但有點不太一樣 @@ 61.219.178.216 07/13
→ eggsu :We consider the following possible "three" cases 04/17 15:16
※ 引述《PttFund (批踢踢基金)》之銘言：
: Let f : R -> R be differentiable and there is a constant c > 0
: such f'(x)≧c for all x. Show that there exists x_0 in R such
: that f(x_0)=0.
自問自答, 先挑簡單的來回答, 其實貼的五題不很簡單.
Proof:
For fixed x in R, we apply MVT to obtain
f(x+1)-f(x) = f'(ξ) ≧ c,
where ξ is between x+1 and x. Thus, f(x+1)≧f(x) + c.
By induction, we have
f(x+n)≧f(x) + nc for positive integers n.
f(x-n)≦f(x) - nc for positive integers n.
We consider the following possible four cases:
(1) f(x) > 0 for all x in R.
(2) f(x) < 0 for all x in R.
(3) f(p)≧0, f(q) < 0 for some p, q in R.
In (3), since f is continuous on R, we can apply intermediate-value
theorem to get the conclusion.
In (1), it is impossible. Fixed x in R. let n > f(x)/c and we have
f(x-n) ≦ f(x) - nc < f(x) - f(x) < 0,
a contradiction. Case (2) is similar.
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