f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证：存在ξ∈（0,1）,使f``(ξ)=2f`(ξ)/1-ξ.f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证：存在ξ∈（0,1）,使f``(ξ)=2f(ξ)/1-ξ.

f(x)在[0,1]上二阶可导,f(0)=0,且f''(x)/f'(x)≠2/(1-x).试证明方程：f（x）/f'(x)=1-x在（0,1）内有且只有一个根 f(x)在(-∞,+∞) 二阶可导,f(x)/x=1,且f''(x)>0,证明f(x)>=x 微积分 设f(x)在[0,1]X上二阶可导,f(1)=f(0)=0设f(x)在[0,1]X上二阶可导,f(1)=f(0)=0,且max f(x)=2 (0 定义在(-1,1)上的函数f(x)满足f(-x)+f(x)=0,且f'(x) 设f(x)在[a,b]上二阶可导,且f''(x)>0,证明:函数F(x)=(f(x)-f(a))/(x-a)在(a,b]上单调增加 设函数f(x)在闭区间[0,1]上可导,且f(0)×f(1) 设f(x)在[0,1]上有二阶连续导数,且满足f(1)=f(0)及|f''(x)| 导数：f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)f(y),且f'(o)=1,求f'(x)f(x+y)=f(x)+f(y)+2xy,且f'(o)存在,求f'(x) f(1+x)=af(x),且f'(0)=b,求f'(1) 设f (x)在x=0处可导,且f (0)=0,求证：lim(x→∞)f (tx)-f (x)/x=(t-1)f' (0) 证明:若函数f(x)在满足关系式f'(x)=f(x),且f(0)=1,则f(x)=e^x 证明：若函数f(x)在(-oo,+oo)内满足关系式f'(x)=f(x),且f(0)=1,则f(x)=e^x 证明:若函数f(x)在满足关系式f'(x)=f(x),且f(0)=1,则f(x)=e^x如题 证明若函数f(x)在R内可导且f'(x)=f(x),f(0)=1,则f(x)=e^x 证明:若函数f(x)在(-∞,+∞)内满足不等式f'(x)=f(x),且f(0)=1,则f(x)=e∧x 设函数f(x)在(-∞,+∞)可导,且满足f(0)=1,f'(x)=f(x),证明f(x)=e^x 高等数学f(x+y)=f(x)+f(y)/1-f(x)f(y),求f(x)f(x+y)=f(x)+f(y)/1-f(x)f(y),则f(x)=tan(ax)怎么证明?f(x)在（-∞,+∞）上有定义,且f'(x)=a(a不等于0) 函数f(x)在定义域R内可导,且f(x)满足 f(x)=f(2-x) (x-1)f'(x)>函数f(x)在定义域R内可导，且f(x)满足 ①f(x)=f(2-x) ②(x-1)f'(x)>0 ③f(3)=0 则不等式xf(x)>0的解集为 设函数f(x)在(-1,1)有定义且满足x≤f(x)≤x²+x证明f'(0)存在且f'(0)=1