1.設函數(shù))
在
處有n階導數(shù),且%3Df''(x_0)%3D%5Ccdots%3Df%5E%7B(n-1)%7D(x_0)%3D0%2Cf%5E%7B(n)%7D%5Cneq%200)
,證明：
(1)當n為奇數(shù)時,f(x)在處不取得極值
(2)當n為偶數(shù)時,f(x)在處取得極值,且當%7D(x_0)%5Clt%200)
時,)
為極大值,當%7D(x_0)%5Cgt%200)
時,為極小值
證：
%E5%9C%A8x_0%E5%A4%84%E5%B8%A6Peano%E4%BD%99%E9%A1%B9%E7%9A%84n%E9%98%B6%E6%B3%B0%E5%8B%92%E5%B1%95%E5%BC%80%E5%BC%8F%E4%B8%BA)
%3Df(x_0)%2B%7Bf%5E%7B(n)%7D(x_0)%5Cover%20n!%7D(x-x_0)%5En%2Bo((x-x_0)%5En))
-f(x_0)%3D%7Bf%5E%7B(n)%7D(x_0)%5Cover%20n!%7D(x-x_0)%5En%2Bo((x-x_0)%5En))
n%E4%B8%BA%E5%A5%87%E6%95%B0%E6%97%B6)
%7D(x_0)%5Cover%20n!%7D(x-x_0)%5En%E5%9C%A8x_0%E4%B8%A4%E4%BE%A7%E5%BC%82%E5%8F%B7)
-f(x_0)%E5%9C%A8x_0%E4%B8%A4%E4%BE%A7%E5%BC%82%E5%8F%B7)
%E5%9C%A8x_0%E5%A4%84%E4%B8%8D%E5%8F%96%E5%BE%97%E6%9E%81%E5%80%BC)
n%E4%B8%BA%E5%81%B6%E6%95%B0%E6%97%B6)
%5En%5Cgt%200%2C%E8%8B%A5%7Bf%5E%7B(n)%7D(x_0)%5Cover%20n!%7D%5Clt%200%2C%E5%88%99f(x)-f(x_0)%5Clt%200%2C%E5%8D%B3f(x_0)%E4%B8%BA%E6%9E%81%E5%A4%A7%E5%80%BC)
%7D(x_0)%5Cover%20n!%7D%5Cgt%200%2C%E5%88%99f(x)-f(x_0)%5Cgt%200%2C%E5%8D%B3f(x_0)%E4%B8%BA%E6%9E%81%E5%B0%8F%E5%80%BC)
2.設常數(shù)
,函數(shù)%3Dlnx-%7Bx%5Cover%20e%7D%2Bk)
在)
內(nèi)有多少零點
解：
%3D%7B1%5Cover%20x%7D-%7B1%5Cover%20e%7D)
%3D0%E5%BE%97%E9%A9%BB%E7%82%B9x%3De)
%5Cgt%200%2Cf(x)%E5%9C%A8(0%2Ce)%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E)
%5Clt%200%2Cf(x)%E5%9C%A8(e%2C%2B%5Cinfty)%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F)
%3D-%5Cinfty%2C%5Clim%5Climits_%7Bx%5Cto%20%2B%5Cinfty%7Df(x)%3D-%5Cinfty%2Cf(e)%3Dk%5Cgt%200)
%E5%9C%A8(0%2Ce)%E5%92%8C(e%2C%2B%5Cinfty)%E5%86%85%E5%9D%87%E6%9C%89%E4%B8%94%E4%BB%85%E6%9C%89%E4%B8%80%E4%B8%AA%E6%A0%B9)
3.設%3Dk)
,求![\lim\limits_{x\to \infty}[f(x+a)-f(x)]](https://math.jianshu.com/math?formula=%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Bf(x%2Ba)-f(x)%5D)
解：
%E5%AD%98%E5%9C%A8)
%E6%98%AF%E8%BF%9E%E7%BB%AD%E5%87%BD%E6%95%B0)
%2C%E6%88%96(x%2Ba%2Cx)%2C%E4%BD%BF%E5%BE%97)
-f(x)%3Df'(%5Cxi)a%2C%E5%85%B6%E4%B8%AD%5Cxi%E4%BB%8B%E4%BA%8Ex%E4%B8%8Ex%2Ba%E4%B9%8B%E9%97%B4)
![\therefore \lim\limits_{x\to \infty}[f(x+a)-f(x)]=\lim\limits_{x\to \infty}af'(\xi)=a\lim\limits_{x\to \infty}f'(\xi)=ak](https://math.jianshu.com/math?formula=%5Ctherefore%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D%5Bf(x%2Ba)-f(x)%5D%3D%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7Daf'(%5Cxi)%3Da%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7Df'(%5Cxi)%3Dak)
4.設
,證明多項式%3Da_0%2Ba_1x%2B%5Ccdots%2Ba_nx%5En)
在(0,1)內(nèi)至少有一個零點
解：
%3Da_0x%2B%7B1%5Cover%202%7Da_1x%5E2%2B%7B1%5Cover%203%7Da_2x%5E3%2B%5Ccdots%2B%7B1%5Cover%20n%2B1%7Da_nx%5E%7Bn%2B1%7D)
![F(x)在[0,1]上連續(xù),在(0,1)內(nèi)可導](https://math.jianshu.com/math?formula=F(x)%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8(0%2C1)%E5%86%85%E5%8F%AF%E5%AF%BC)
%3D0%2CF(1)%3Da_0%2B%7Ba_1%5Cover%202%7D%2B%5Ccdots%2B%7Ba_n%5Cover%20n%2B1%7D%3D0)
%2C%E4%BD%BFF'(%5Cxi)%3D0)
%3Df(x)%3Da_0%2Ba_1x%2B%5Ccdots%2Ba_nx%5En)
%E5%9C%A8(0%2C1)%E5%86%85%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%80%E4%B8%AA%E9%9B%B6%E7%82%B9)
5.設f(x)在[0,a]上連續(xù),在(0,a)內(nèi)可導,且f(a)=0,證明存在一點)
,使%2B%5Cxi%20f'(%5Cxi)%3D0)
解：
![設F(x)=xf(x),則F(x)在[0,a]上連續(xù),在(0,a)內(nèi)可導](https://math.jianshu.com/math?formula=%E8%AE%BEF(x)%3Dxf(x)%2C%E5%88%99F(x)%E5%9C%A8%5B0%2Ca%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8(0%2Ca)%E5%86%85%E5%8F%AF%E5%AF%BC)
%3DF(a)%3D0)
%2C%E4%BD%BFF'(%5Cxi)%3D0)
%2B%5Cxi%20f'(%5Cxi)%3D0)
6.
解：
(x{x-1})'=[e{(x-1)lnx}]'=x^{x-1}(lnx+{x-1\over x})$
%5Cover%20%7B1-(lnx%2B1)%5Cover%20x%5E2%7D-1%7D)
)
%3D2)
7.![\lim\limits_{x\to 0}[{1\over ln(1+x)}-{1\over x}]](https://math.jianshu.com/math?formula=%5Clim%5Climits_%7Bx%5Cto%200%7D%5B%7B1%5Cover%20ln(1%2Bx)%7D-%7B1%5Cover%20x%7D%5D)
解：
![原式=\lim\limits_{x\to 0}[{x-ln(1+x)\over xln(1+x)}]](https://math.jianshu.com/math?formula=%E5%8E%9F%E5%BC%8F%3D%5Clim%5Climits_%7Bx%5Cto%200%7D%5B%7Bx-ln(1%2Bx)%5Cover%20xln(1%2Bx)%7D%5D)
![=\lim\limits_{x\to 0}[{x-ln(1+x)\over x^2}]](https://math.jianshu.com/math?formula=%3D%5Clim%5Climits_%7Bx%5Cto%200%7D%5B%7Bx-ln(1%2Bx)%5Cover%20x%5E2%7D%5D)
![=\lim\limits_{x\to 0}[{1-{1\over 1+x}\over 2x}]](https://math.jianshu.com/math?formula=%3D%5Clim%5Climits_%7Bx%5Cto%200%7D%5B%7B1-%7B1%5Cover%201%2Bx%7D%5Cover%202x%7D%5D)
![=\lim\limits_{x\to 0}[{1\over 2(1+x)}]={1\over 2}](https://math.jianshu.com/math?formula=%3D%5Clim%5Climits_%7Bx%5Cto%200%7D%5B%7B1%5Cover%202(1%2Bx)%7D%5D%3D%7B1%5Cover%202%7D)
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