If f (x) = 1 of a + (2 ^ x + 1) is an odd function, then a =?

If f (x) = 1 of a + (2 ^ x + 1) is an odd function, then a =?

First consider whether x = 0 is in the domain
If f (0) = 0, then f (0) = 0
a=-1/2
Reexamination
If not,
Then f (x) + F (- x) = 0
The solution is a = - 1 / 2
The result is the same

Exponential function in senior one
Calculation √ (E + e ^ - 1) ^ 2-4 + √ (E-E ^ - 1) ^ 2 + 4 (E ≈ 2.7)
PS: √ (E + e ^ - 1) ^ 2-4, and √ (E-E ^ - 1) ^ 2 + 4 are two parts
Kneel down! Kneel down!

√（e+e^(-1）)^2-4 =√e^2+2+e^(-2)-4
=√e^2-2+e^(-2)=√(e-e^(-1))^2=e-e^(-1)
√(e-e^(-1))^2+4=√e^2-2+e^(-2)+4
=√e^2+2+e^(-2)=√(e+e(-1))^2=e+e^(-1)
√（e+e^-1）^2-4 + √(e-e^-1)^2+4=2*e

Exponential function problem~
Find the range y = 2 ^ X-1 / 2 ^ x + 1

It's easy to make y = 1 - (2 / 2 ^ x + 1)
Where 2 ^ x + 1 > 1
Then 2 / 2 ^ x + 10
The final result is (- 1,1)
(it hasn't been inspected. Please check it yourself.)