设f(x)=2^(x-1)+1/2^(x+1),证明f(x+a)+f(x-a)=2f(x)f(a)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/27 03:28:32
设f(x)=2^(x-1)+1/2^(x+1),证明f(x+a)+f(x-a)=2f(x)f(a)
f(x)=2^(x-1)+1/2^(x+1)
f(x+a)=2^(x+a-1)+1/2^(x+a+1)
f(x-a)=2^(x-a-1)+1/2^(x-a+1)
f(x)f(a)
=[2^(x-1)+1/2^(x+1)][2^(a-1)+1/2^(a+1)]
=2^(x+a-2)+2^(x-1-a-1)+1/2^(x+1-a+1)+1/2^(x+1+a+1)
=(1/2)[2^(x+a-1)+2^(x-a-1)+1/2^(x-a+1)+1/2^(x+a+1)]
=(1/2)[f(x+a)+f(x-a)]
f(x+a)+f(x-a)=2f(x)f(a)
f(x+a)=2^(x+a-1)+1/2^(x+a+1)
f(x-a)=2^(x-a-1)+1/2^(x-a+1)
f(x)f(a)
=[2^(x-1)+1/2^(x+1)][2^(a-1)+1/2^(a+1)]
=2^(x+a-2)+2^(x-1-a-1)+1/2^(x+1-a+1)+1/2^(x+1+a+1)
=(1/2)[2^(x+a-1)+2^(x-a-1)+1/2^(x-a+1)+1/2^(x+a+1)]
=(1/2)[f(x+a)+f(x-a)]
f(x+a)+f(x-a)=2f(x)f(a)
设函数f(x)满足f(x)+2f(1/x)=x,求f(x)
设f(x)=(a^x+a^y) (a>0),证明f(x+y)+f(x-y)=2f(x)f(y)
f(x)在x=a处有二阶导数,求证x趋于0时lim(((f(a+x)-f(a)/x}-f‘(a))/x=1/2f''(a
设函数f(x)是二次多项式,证明f(x)=f ''(a)/2*(x-a)^2+f '(a)(x-a)+f(a)
设映射f:X——Y,A包含于X,B包含于X,证明1,f(A并B)=f(A)并f(B) 2,f(A交B)包含于f(A)交f
设函数f(x)连续 (1)证明:∫上a下-af(x)dx=1/2∫上a下-a[f(x)+f(-x)
证明一个函数的周期设a>0,如果f(x)+f(x+a)+f(x+2a)+f(x+3a)+f(x+4a)=f(x)f(x+
已知f(x)=a-[2/(2^x+1)] 且 f(-x)=-f(x),(1)求f(x)的值域(2)f(x)的反函数f^-
设函数f(x)=a-2/2^x+1
已知f(x)=x(1/(2^x-1)+1/2).(1)证明f(x)大于0.(2)设F(x)=f(x+t)-f(x-t).
设f(x)=lg(2/(1-x)+a)是奇函数,则使f(x)
1.设F(X)=lg[2/(1-x)+a]是奇函数,即使 F(X)