∵f(x)=ex-e-x+1,
f(a)=2,
∴ea-e-a+1=2,
∴ea-e-a=1,
∴f(-a)=e-a-ea+1=-(ea-e-a)+1=-1+1=0.
故选D.
已知函数f(x)=ex-e-x+1(e是自然对数的底数),若f(a)=2,则f(-a)的值为( ) A.3 B.2 C.1 D.0
已知函数f(x)=ex-e-x+1(e是自然对数的底数),若f(a)=2,则f(-a)的值为( )
A. 3
B. 2
C. 1
D. 0
A. 3
B. 2
C. 1
D. 0
数学人气:287 ℃时间:2019-08-20 21:24:28
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