For the function f : ℝ -> ℝ
it's true that : f(x + y) = f(x)*f(y) , for every x, y∈ℝ and that f(0) ≠ 0
Prove that :
a) f(x) ≠ 0 , for every x∈ℝ
b) f(x) > 0 , for every x∈ℝ
c) f(0) = 1
d) f(x)*f(-x) = 1 , for every x∈ℝ
e) If the function f(x) = 1 has a single solution, then the f can be inversed and it's true that
[tex]f ^{ - 1} (xy) = f^{ - 1}(x) \: + \: f^{ - 1}(y) [/tex]
for every x , y > 0
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