设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0是c^2f(c)+2cf'(c)=0,我确定希望那位大师帮帮忙
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/17 05:36:35
设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0是c^2f(c)+2cf''(c)=0,我确定希望那位大师帮帮忙设f(x
设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0是c^2f(c)+2cf'(c)=0,我确定希望那位大师帮帮忙
设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0
是c^2f(c)+2cf'(c)=0,我确定希望那位大师帮帮忙
设f(x)在[0,a]连续,在(0,a)内可导,且f(a)=0,证明;存在一点c属于(0,a),使c^2f(c)+2cf(c)=0是c^2f(c)+2cf'(c)=0,我确定希望那位大师帮帮忙
应该是c²f'(c)+2cf(c) = 0吧.
设g(x) = x²f(x),则g(c)在[0,a]连续,在(0,a)可导,且g(0) = 0 = g(a).
由罗尔定理,存在c∈(0,a)使g'(c) = 0,即有c²f'(c)+2cf(c) = 0.
不可能是c²f(c)+2cf'(c) = 0.
反例如f(x) = (x-a)·e^(-x²/4),易验证满足题目条件.
f'(x) = -x/2·e^(-x²/4)·(x-a)+e^(-x²/4) = (-x²+ax+2)/2·e^(-x²/4).
2xf'(x) = (-x³+ax²+2x)·e^(-x²/4).
x²f(x) = (x³-ax²)·e^(-x²/4).
则x²f(x)+2xf'(x) = 2x·e^(-x²/4) > 0对任意x > 0.
即不存在c > 0使c²f(c)+2cf'(c) = 0.
好像忘记了
证明:设f(x)在[a,b]上连续,在(a,b)内可导,(0
设f(x)在[a,b]上连续,在(a,b)上可导(0
设函数f(x)在[a,b]上连续,在(a,b)内可导(0
设f(x)在[a,b]上连续,在(a,b)内可导,(0
设f(x)在[a,b]上连续,在(a,b)内可导(0
设f(x)在[a,b]上连续,在(a,b)内可导(0
设f(x)在[a,b]上连续,且a
设函数f(x)在[a,b]上连续,a
设f(x)在[a,b]上连续,且a
设f(x)在[a,b]上连续,且a
设f(x)在[a,b]上连续,a
设函数f(x)在[a,b]上连续,a
设函数f(x)在对称区间【-a,a】上连续,证明∫(-a,a)f(x)dx=∫(0,a)[f(x)+f(-x)]dx
【50分高数微积分题】设f(x)在[a,b]上连续,在(a,b)内可导 f(a)f(b)>0 f(a)f[(a+b)/2]
设f(x)在[a,b]上连续,在(a,b)内可导,f(a)f(b)>0,f(a)f[(a+b)/2]
设函数f(x)在[a,b]上连续,在(a,b)可导,且f(a)*f(b)>0,f(a)*f((a+b)/2)
求设f'(x)在[0,a]上连续.f(0)=0,证明|定积分f(x)d(x)
设F(x)=(f(x)-f(a))/(x-a),(x>a)其中f(x)在[a,+∞)上连续,f''(x)在(a,+∞)内存在且大于0,求证F(x)在(a,+∞)内单调递增.