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展开全部解：（法一）直接用公式f'(x)=f(x)+1即 f'(x)-f(x)=1故f(x)=e^∫dx [ ∫ e^∫-dx dx +c]=e^x[-e^(-x)+c]=ce^x-1又f(0)=0得c=1故f(x)=e^x-1（法二）dy/dx=y+1故dy/(y+1)=dx积分ln|y+1|=x+lncy+1=ce^x又f(0)=0得c=1故f(x)=e^x-1已赞过已踩过你对这个回答的评价是？评论
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0:-1;if(n!==r)for(var i=0;i0&&function(t,e,n,r){var i=document.getElementsByClassName(t);if(i.length>0)for(var o=0;o展开全部dy/dx=y+1dy/(y+1)=dx两边积分得通解为ln|y+1|=x+C
y+1|=e^(x+c)即f(x)=-1±e^(x+C)∵f(0)=0∴取+号，0=-1+e^C,c=0∴f(x)=-1+e^x
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（1）求证：f(x)是偶函数（2）求证：f(x)是周期函数（3）若f(x)在[0，1]内是单调函数，求f(1/3)与f(1/6)的值高中数学会的说下谢谢！...
（1）求证：f(x)是偶函数（2）求证：f(x)是周期函数（3）若f(x)在[0，1]内是单调函数，求f(1/3)与f(1/6)的值高中数学 会的说下 谢谢！
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（1）将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0)，即2f(0)=2f(0)f(0)，由f(0)≠0得,f(0)=1, 再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得, f(y)+f(-y)=2f(y), f(-y)=f(y),故f(x)是偶函数。（2）将y=1/2代入f(x+y)+f(x-y)=2f(x)f(y)得f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0.由f(1/2)=0 得 f(x+1/2)+f(x-1/2) =0.f(x+1/2)=-f(x-1/2) , 同理f(x-1/2)= f(x-1+1/2)=- f(x-1-1/2)= - f(x-3/2)，故f(x+1/2)= -f(x-1/2) =f(x-3/2), 设y= x-3/2,则x=y+3/2,x+1/2=y+2,代入上式得f(y+2)=f(y),故f(x)是周期为2的周期函数。（3）设x=y代入f(x+y)+f(x-y)=2f(x)f(y)得f(2x)+f(0)=2f(x)f(x),f(2x) =2f(x)f(x)-1, 将x=1/2代入得f(1) =2f(1/2)f(1/2)-1=-1,即f(1) =-1，设y=2x代入f(x+y)+f(x-y)=2f(x)f(y)得,f(3x)+f(-x)=2f(x)f(2x),故f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)将x=1/3代入上式得f(1)=4 f(1/3) f(1/3) f(1/3)-3 f(1/3)，设a=f(1/3)得4a^3-3a+1=0,解得a=1/2，即f(1/3)=1/2，再x=1/6代入f(2x) =2f(x)f(x)-1得f(1/3) =2f(1/6)f(1/6)-1，设b=f(1/3)得2bb-1= f(1/3)=1/2, 4bb-3=0,b=√3/2. f(x)在[0,1]内单调t得f(1/6)=√3/2。
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收起(1)令y=0,等式变为:f(x)+f(x)=2f(x)f(0),所以f(0)=1;再x=0,等式化为:f(y)+f(-y)=2f(0)f(y),所以f(y)=f(-y),偶函数得证;(2)令y=1/2,等式化为:f(x+1/2)+f(x-1/2)=0;再x-1/2=z,则f(z)=-f(z+1)=f(z+2),周期函数,最小正周期T=2;(3)令x=1/3;y=1/6,等式为:f(1/3+1/6)+f(1/3-1/6)=2f(1/3)f(1/6),即f(1/6)[2f(1/3)-1]=0.因为单调,且f(0)=1＞f(1/2)=0,即单调减函数,所以f(1/6)＞f(1/3)＞f(1/2)=0;f(1/3)=1/2;令x=y=1/3,等式为:f(1/6+1/6)+f(1/6-1/6)=2f(1/6)f(1/6),f(1/6)=√3/2.（1）将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0)，即2f(0)=2f(0)f(0)，由f(0)≠0得,f(0)=1, 再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得, f(y)+f(-y)=2f(y), f(-y)=f(y),故f(x)是偶函数。（2）将y=1/2代入f(x+y)+f(x-y)=2f(x)f(y)得f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0.由f(1/2)=0 得 f(x+1/2)+f(x-1/2) =0.f(x+1/2)=-f(x-1/2) , 同理f(x-1/2)= f(x-1+1/2)=- f(x-1-1/2)= - f(x-3/2)，故f(x+1/2)= -f(x-1/2) =f(x-3/2), 设y= x-3/2,则x=y+3/2,x+1/2=y+2,代入上式得f(y+2)=f(y),故f(x)是周期为2的周期函数。（3）设x=y代入f(x+y)+f(x-y)=2f(x)f(y)得f(2x)+f(0)=2f(x)f(x),f(2x) =2f(x)f(x)-1, 将x=1/2代入得f(1) =2f(1/2)f(1/2)-1=-1,即f(1) =-1，设y=2x代入f(x+y)+f(x-y)=2f(x)f(y)得,f(3x)+f(-x)=2f(x)f(2x),故f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)将x=1/3代入上式得f(1)=4 f(1/3) f(1/3) f(1/3)-3 f(1/3)，设a=f(1/3)得4a^3-3a+1=0,解得a=1/2，即f(1/3)=1/2，再x=1/6代入f(2x) =2f(x)f(x)-1得f(1/3) =2f(1/6)f(1/6)-1，设b=f(1/3)得2bb-1= f(1/3)=1/2, 4bb-3=0,b=√3/2. f(x)在[0,1]内单调t得f(1/6)=√3/2。 所以f(1/3)=1/2
f(1/6)=3/2这个我还真的忘了，都大学毕业工作几年了}