f(x)= 12π√e−(x−1)22
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(1)由题意可得f(x)的定义域为(0,+∞),f′(x)=(x−1)[x−(a−1)]x2∵a<2,∴a-1<1①当a-1≤0,即a≤1,∴x∈(0,1)时,f′(x)<0,f(x)是减函数,x∈(
如图所示.
∵f(x)是奇函数且f(x+2)=-f(x),∴f(x+4)=-f(x+2)=f(x)∴函数f(x)的周期T=4.∵当0≤x≤1时,f(x)=12x,又f(x)是奇函数,∴当-1≤x≤0时,f(x)=
∵ex在(-∞,1)上单调递增,ex-2在(-∞,1)上单调递增∴函数f(x)=12(ex+ex−2)在(-∞,1)上单调递增∴函数f(x)=12(ex+ex−2)的值域为(0,12(e+1e))则反
∫f'(x)dx/1+f^2(x)=∫df(x)/[1+f^2(x)]=arctanf(x)+c=arctan(e^x/x)+c
复合函数求导:f'(x)=[e^(-x)*ln(2-x)]'+[(1+3x^2)^(1/2)]'=[e^(-x)]'*ln(2-x)+e^(-x)*[ln(2-x)]'+[(1+3x^2)^(1/2)
令e^x=u,则dx=du/u原式=∫(u³+u)/(u(u^4-u²+1))du=∫(u²+1)/(u^4-u²+1)du=∫(1+1/u²)/(u
∵函数f(x)=12(x−1)2+1的定义域和值域都是[1,b],且f(x)在[1,b]上为增函数,∴当x=1时,f(x)=1,当x=b时,f(x)=12(b-1)2+1=b,解得:b=3或b=1(舍
(1)∵函数f(x)=−12+12x+1,∴f(−x)=−12+12−x+1=−12+2x1+2x….(2分)=−12+1−11+2x=12−11+2x=−f(x)….(4分)又函数f(x)的定义域为
1.f'(x)=e^x-1/(x+1),f'(0)=0,f''(x)=e^x+1/(x+1)^2>0,f'(x)为(-1,+∞)上的增函数,所以x>0时,f'(x)>f'(0)=0,f(x)在(0,+
f(x)=(e^x+1)/e^x=1+1/e^x=1+e^(-x)f'(x)=[1+e^(-x)]'=[e^(-x)]'=[e^(-x)]*(-x)'=[e^(-x)]*(-1)=-e^(-x)=-1
∵f(x)是偶函数,∴f(-1)=f(1),∴u=0∴f(x)=e−x2,∴当x=0时函数f(x)取得最大值,且最大值为1,∴m+μ=1.故答案为:1.
(1)令x−1=t,则t≥-1,x=t+1,x=(t+1)2.∴f(t)=(t+1)2+2(t+1)+2,即f(t)=t2+4t+5.把t换成x得f(x)=x2+4x+5.(2)由(1)可知:x−1=
(I)∵函数f(x)=axlnx−bx(x>0,x≠1),∴f′(x)=−a(1+lnx)(xlnx)2+bx2,∵f(x)在x=e处的切线与x轴平行,∴f′(e)=0,即−a(1+lne)(elne
若f(x)=e^x/(1+e^x)+x∫(0→1)f(x)dx求f(x)对f(x)=e^x/(1+e^x)+x∫(0→1)f(x)dx两边积分得∫(0→1)f(x)dx=∫(0→1)[e^x/(1+e
f'(x)=(e^x)'-x'-1'=e^x-1-0=e^x-1
f'(x)=[(1+x)/(1-x)]'e^(-ax)+(-ae^-ax)[(1+x)/(1-x)]=[(1+x)/(1-x)]'e^(-ax)-ae^(-ax)*(1+x)/(1-x)=[-(x-1
=f[e^x/(1+e^2x)]dx=f[1/(1+e^2x)]de^x=arctan(e^x)
一般的[f(x)/g(x)]'=[f'(x)g(x)-f(x)g'(x)]/[g^2(x)]所以对本题目f'(x)=[e^x*(x-1)-e^x*1]/(x-1)^2=e^x*(x-2)/(x-1)^