设函数y=f(x)在点x=0的某个邻域内有连续的二阶导数,且f'(0)=0,lim(x->0) ,则( ). A.f(0)是f(x)的极大值B.f(0)是f(x)的极小值C.(0,f(0))是曲线y=f(x)的拐点D.f(0)不是f(x)的极值,(0,f(0))也不是曲线y=f(x)的拐点...
设函数f(x)在x=0的某邻域内具有二阶连续导数,且,则A.f(0)是f(x)的极大值.B.f(0)是f(x)的极小值.C.(0,f(0))是曲线f(x)的拐点.D.f(
设函数f(x)在x=0的某邻域内有连续的二阶导数,且 f''(0)=f'(0)=0 ,则(A)x=0是f(x)的零点(B)x=0为f(x)极小值点(C)当lim_(x→0)(f'(x))/(|x|)=1 时,(0,f(0)为拐点:(D)当lim_(x→0)(f''(x))/(sinx)=1时,(0,f(0)为拐点 ...
二阶麦克劳林公式为:f(x)=f(0)+f'(0)x+(f″(0))(2!)x^2+o(x^2) 故:af(h)+bf(2h)+cf(3h)-f(0)=(a+b+c-1)f(0)+f'(0)⋅ (ah+2bh+3ch)+f″(0)⋅ (ah^2+4bh^2+9ch^2)2+o(h^2)=o(h^2); f(0)、f'(0)、f″(0)≠q 0,h为自变量,所以有: \((...
【解析】证由条件 lim_(x→0)(f(x))/x=0 可得 f(0)=f'(0)=0 由于lim_(x→0)(f(x))/(x^2)=lim_(x→0)f'(x)=lim_(x→0)/(f'(x))/2=(f'(0))/2 故lim_(x→0) rac(|x| rac1( rac1(x^2))= rac(1f'^2)2 由于∑_(n=1)^∞1/(n^2)收敛,故∑_(n=1)^∞|f...
百度试题 题目设函数f(x)在点x=0的某邻域内具有连续的二阶导数,且f"(0)=f'(0)=0,则() A.点x=0为f(x)的零点B.点x=0为f(x)的极值点C.当时,(0,f(0))为拐点D.当时,(0,f(0))为拐点相关知识点: 试题来源: 解析 D 反馈 收藏
【题目】设函数f(x)在x=0的某邻域内有连续的二阶导数,且 f''(0)=f'(0)=0 ,则(A)x=0是f(x)的零点(B)x=0为f(x)极小值点)当 lim_(
设函数f(x)在x=0的某邻域内具有二阶连续导数。且f(0)≠0,f(0)≠0,f'(0)≠0.证明:存在唯一的一组实数λ1,λ2,λ3,使得当h→0时,λ1f(h)+λ
【题目】设函数f(x)在点x=0的某邻域内有连续的二阶导数,且 f'(0)=f''(0)=0 ,则(A)f(0)=0(B)点(0,f(0))是曲线y=f(x)的拐点(C)当lim_(x→0)(f'(x))/(cosx)=1 时,(0,f(0)是拐点(D)当lim_(x→0)f''(x))(/sinx)=1时,(0,f(0)是拐点 ...
设函数f(x)在x=0处的某邻域内有二阶连续导数,且f(0)不为0,f'(0)不为0,f''(0)不为0,.转下面 证明,存在唯一的一组实数abc,使得当h趋向于0时