f(x)在[0,1]上具有二阶导数,|f(x)|
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 15:13:38
f(x)在[0,1]上具有二阶导数,|f(x)|f(x)在[0,1]上具有二阶导数,|f(x)|f(x)在[0,1]上具有二阶导数,|f(x)|证明:把f(x)在x=c处泰勒展开得f(0)=f(c)-
f(x)在[0,1]上具有二阶导数,|f(x)|
f(x)在[0,1]上具有二阶导数,|f(x)|
f(x)在[0,1]上具有二阶导数,|f(x)|
证明:把f(x)在x=c处泰勒展开得
f(0)=f(c)-f'(c)*c+f''(m)*c²/2
f(1)=f(c)+f'(c)*(1-c)+f''(n)*(1-c)²/2
两式相减,得
f(1)-f(0)=f`(c)+f''(n)*(1-c)²/2-f''(m)*c²/2
f'(c)=f(1)-f(0)+f''(m)*c²/2-f''(n)*(1-c)²/2
所以
|f'(c)|≤|f(1)|+|f(0)|+|f''(m)|*c²/2+|f''(n)|*(1-c)²/2
因为|f(x)|≤a,所以|f(0)≤a |f(1)|≤a,同样|f``(x)|≤b,所以|f``(m)|≤b | f``(n)|≤b
所以f(c)≤a+a+b/2*[c²+(1-c)²]
又0
设f(x)在[0,1]上具有二阶连续导数,且|f''(x)|
f(x)在[0,1]上具有二阶导数,|f(x)|
设f(x)在(0,1)具有二阶导数,且|f(x)|
积分应用 设f (x)在[0,1]上具有二阶连续导数,若f ( π ) = 2,∫ [ f (x)+ f (x)的二阶导数]sin xdx =5,求f (0) ..
设f(x)具有二阶导数,g(x)=f(0)(1-x)+f(1)x,则在区间[0,1]上
一道高数题设函数f(x)在〔o,1〕上具有二阶导数,具满足条件|f(x)|
已知f(x)在【0,1】上具有二阶导数且f(0)=f(1)=0设F(x)=xf(x)证明:在(0,1)内方程F’’(x)=0存在实数根
f(x)在(a,b)上具有二阶连续导数又 f'(a)=f'(b)=0 证明:存在u属于(a,b) f(u)
设f(x)在(0,1)上具有二阶连续导数,若f(π)=2,∫ (0到π)[f(x)+f(x)]sinxdx=5,求f(0)
f(x)在(a,b)上具有二阶连续导数又 f'(a)=f'(b)=0 证明:存在u属于(a,b) f(u)f(x)在[a,b]上具有二阶连续导数又 f'(a)=f'(b)=0 证明:存在u属于(a,b),| f''(u)|>=4|f(a)-f(b)|/(b-a)^2
设函数f(x)在闭区间[a,b]上具有二阶导数,且f(x)>0,证明∫(a,b)f(x)dx>f(设函数f(x)在闭区间[a,b]上具有二阶导数,且f(x)>0,证明∫(a,b)f(x)dx>f(a+b/2)(b-a)
f(x)在点x=0处具有连续的二阶导数,证明f证明f(x)的二阶导数有界
设函数f(x)在[0,1]上具有二阶导数,且f(0)=f(1)=0,minf(x)=—1 x∈[0.1].证明maxf''(x)大于等于8
f(x)在(0,+∞)上具有二阶导数,对一切x>0有|f(x)|≤a,|f''(x)|≤b,a,b为常数.证明:|f'(x)|≤2√ab
f(x)在(0,+∞)上具有二阶导数,对一切x>0有|f(x)|≤a,|f''(x)|≤b,a,b为常数.证明:|f'(x)|≤2√ab
设函数f(x)在x=0处具有二阶导数,且f(0)=0,f’(0)=1,f’’(0)=3,求极限lim(x->0)(f(x)-x)/x^2
设函数f(x)在x=0处具有二阶导数,且f(0)=0,f’(0)=1,f’’(0)=3,求极限lim(x->0)(f(x)-x)/x^2
设f(x)在[a,b]上具有二阶导数 且f(a)=f(b)=0 f'(a)f'(b)>0 证明 至少存在一点设f(x)在[a,b]上具有二阶导数 且f(a)=f(b)=0 f'(a)f'(b)>0 证明 至少存在一点c属于(a,b),使f‘’(c)=0