1.lim [(ln(1+x))/(x^3)+f(x)/(x^2)]=0x->0求f''(0).2.f(x)在(0,1)上连续且有二阶导数,f(0)=f(1)=0,在0到1区间上有f(x)max=2.求证存在u在0~1内且f''(u)第二题明白了第一题没看懂,而且你泰勒展错了第一题无法L'Hos
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![1.lim [(ln(1+x))/(x^3)+f(x)/(x^2)]=0x->0求f''(0).2.f(x)在(0,1)上连续且有二阶导数,f(0)=f(1)=0,在0到1区间上有f(x)max=2.求证存在u在0~1内且f''(u)第二题明白了第一题没看懂,而且你泰勒展错了第一题无法L'Hos](/uploads/image/z/14788270-46-0.jpg?t=1.lim+%5B%28ln%281%2Bx%29%29%2F%28x%5E3%29%2Bf%28x%29%2F%28x%5E2%29%5D%3D0x-%3E0%E6%B1%82f%27%27%280%29.2.f%28x%29%E5%9C%A8%280%2C1%29%E4%B8%8A%E8%BF%9E%E7%BB%AD%E4%B8%94%E6%9C%89%E4%BA%8C%E9%98%B6%E5%AF%BC%E6%95%B0%2Cf%280%29%3Df%281%29%3D0%2C%E5%9C%A80%E5%88%B01%E5%8C%BA%E9%97%B4%E4%B8%8A%E6%9C%89f%28x%29max%3D2.%E6%B1%82%E8%AF%81%E5%AD%98%E5%9C%A8u%E5%9C%A80%EF%BD%9E1%E5%86%85%E4%B8%94f%27%27%28u%29%E7%AC%AC%E4%BA%8C%E9%A2%98%E6%98%8E%E7%99%BD%E4%BA%86%E7%AC%AC%E4%B8%80%E9%A2%98%E6%B2%A1%E7%9C%8B%E6%87%82%EF%BC%8C%E8%80%8C%E4%B8%94%E4%BD%A0%E6%B3%B0%E5%8B%92%E5%B1%95%E9%94%99%E4%BA%86%E7%AC%AC%E4%B8%80%E9%A2%98%E6%97%A0%E6%B3%95L%27Hos)
1.lim [(ln(1+x))/(x^3)+f(x)/(x^2)]=0x->0求f''(0).2.f(x)在(0,1)上连续且有二阶导数,f(0)=f(1)=0,在0到1区间上有f(x)max=2.求证存在u在0~1内且f''(u)第二题明白了第一题没看懂,而且你泰勒展错了第一题无法L'Hos
1.lim [(ln(1+x))/(x^3)+f(x)/(x^2)]=0
x->0
求f''(0).
2.f(x)在(0,1)上连续且有二阶导数,f(0)=f(1)=0,在0到1区间上有f(x)max=2.求证存在u在0~1内且f''(u)
第二题明白了
第一题没看懂,而且你泰勒展错了
第一题无法L'Hospital下去,我忘了说了,题目只说有二阶导数,三阶未必存在,而求f''(0)好像必须用到三阶
(0阶1阶我就是用1阶2阶算出的)
1.lim [(ln(1+x))/(x^3)+f(x)/(x^2)]=0x->0求f''(0).2.f(x)在(0,1)上连续且有二阶导数,f(0)=f(1)=0,在0到1区间上有f(x)max=2.求证存在u在0~1内且f''(u)第二题明白了第一题没看懂,而且你泰勒展错了第一题无法L'Hos
1.ln(1+x)=x+x^2/2+x^3/3+x^4/4+……
所以ln(1+x)/x=1/x^2+1/(2x)0+1/3+……
而f(x)=f(0)+f'(0)x+f''(0)x^2/2!+……
f(x)/x^2=……+f''(0)/2+……
说明:f''(0)=-2/3.
2.最大值点为e.则f(e)=2;f'(e)=0.
f(0)=f(e)+f''(w1)(x-e)^2/2=0
f(1)=f(e)+f''(w2)(x-e)^2/2=0
即2+f''(w1)*e^2/2=0
2+f''(w2)*(1-e)^2/2=0
当0
1:f''(0)=-2/3。
2:自己做或见我的博客