帮帮忙，各位大神
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f(x)= a - 2/(2^x+1)
令x1＜x2
f(x2)-f(x1)= [ a - 2/(2^x2+1) ] - [ a - 2/(2^x1+1) ]
= - 2/(2^x2+1) + 2/ (2^x1+1)
= (2^x2+1-2^x1-1)/[(2^x2+1)(2^x1+1)]
= (2^x2-2^x1)/[(2^x2+1)(2^x1+1)]
∵x2＞x1，∴2^x2＞2^x1，∴2^x2-2^x1＞0
又：[(2^x2+1)(2^x1+1)]＞0
∴f(x2)-f(x1)＞0
即：f(x2)＞f(x1) ，得证。
f（x）是奇函数，则f(-x) = -f(x)
a - 2/(2^(-x)+1) = -a + 2/(2^x+1)
2a = 2/(2^(-x)+1) + 2/(2^x+1)
2a = 2*2^x /(1+2^x) + 2/(2^x+1) = 2*(2^x+1)/(2^x+1) = 2
a = 1
令x1＜x2
f(x2)-f(x1)= [ a - 2/(2^x2+1) ] - [ a - 2/(2^x1+1) ]
= - 2/(2^x2+1) + 2/ (2^x1+1)
= (2^x2+1-2^x1-1)/[(2^x2+1)(2^x1+1)]
= (2^x2-2^x1)/[(2^x2+1)(2^x1+1)]
∵x2＞x1，∴2^x2＞2^x1，∴2^x2-2^x1＞0
又：[(2^x2+1)(2^x1+1)]＞0
∴f(x2)-f(x1)＞0
即：f(x2)＞f(x1) ，得证。
f（x）是奇函数，则f(-x) = -f(x)
a - 2/(2^(-x)+1) = -a + 2/(2^x+1)
2a = 2/(2^(-x)+1) + 2/(2^x+1)
2a = 2*2^x /(1+2^x) + 2/(2^x+1) = 2*(2^x+1)/(2^x+1) = 2
a = 1
追答
对于任何实数a，f（x）是增函数
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