设f在[0,+]上连续,满足证明:(1){an}为收敛数列;(2)设(3)若条件改为
设f在[0,+]上连续,满足
证明:
(1){an}为收敛数列;
(2)设
(3)若条件改为
设f在[0,+]上连续,满足
证明:
(1){an}为收敛数列;
(2)设
(3)若条件改为
设函数f(x),F(x)在[a,b]上连续,在(a,b)内可导,且F'(x)≠0,x∈(a,b).由于f(x),F(x)在[a,b]上都满足拉格朗日中值定理的条件,故存在点ξ∈(a,b),使
f(b)-f(a)=f'(ξ)(b-a), (1)
F(b)-F(a)=F'(ξ)(b-a), (2)
又,F'(x)≠0,x∈(a,b),(1),(2)两式相除,即有
,
以上证明柯西中值定理的方法对吗?
设f(x)在[0,+∞)上连续,在(0,+∞)内可导且满足
f(0)=0,f(x)≥0,f(x)≥f'(x)(x>0),
证明:f(x)≡0.
设f(x)在[0,+∞)上连续,在(0,+∞)内可导且满足
f(0)=0,f(x)≥0,f(x)≥f'(x)(),证明:f(x)=0.
设函数f(x)在[a,b]上连续,且满足f(a)=f(b)=0,f'+(a),f'-(b)存在,f'+(a)·f'-(b)>0证明:f(x)在(a,b)内存在零点
设f(x)在[0,+∞)上连续,在(0,+∞)内可导且满足
f(0)=0, f(x)≥0,f(x)≥f'(x)(x>0),
证明:f(x)0.
设函数f(x)在[0,1]上连续,在(0,1)内可微,且满足证明在(0,1)内至少有一点a,使f(a)+af'(a)=0
(1)设f(x)在[0,+∞)上连续,可导,且证明:存在c∈(0,+∞),使
f'(c)=0
设f(x)在[0,1]上连续,在(0,1)内可导,且满足
证明至少存在一点ξ∈(0,1),使得f'(ξ)=(1-ξ-1)f(ξ)
设f(x)在[0,1]上连续,且0≤f(x)≤1,证明:至少存在一点ζ∈[0,1] ,使f(ζ)=ζ .
设函数f(x)在(0.+∞)上满足方程
证明:f(x)=f(1),x∈(0,+∞).