f(x)=log3(3^x+1)+1/2ax是偶函数,则a=

问题描述:

f(x)=log3(3^x+1)+1/2ax是偶函数,则a=

f(x)=f(-x)
f(x)-f(-x)=0
log3(3^x+1)+1/2ax-log3(3^-x+1)+1/2ax=0
log3[(3^x+1)/(3^-x+1)]+ax=0
log3[3^(x+1+x-1)]+ax=0
log3[3^(2x)]+ax=0
2x+ax=0
(2+a)x=0
这是恒等式则2+a=0
a=-2