Question

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

`f ^( - 1) (xy) = f^( - 1)(x) \: + \: f^( - 1)(y)`
for every x , y > 0
​

Answer 1

a) f(x) ≠ 0, for every x∈ℝ

Step-by-step explanation:

Since f(0) ≠ 0, it follows that f(0) is either positive or negative. Let's consider the case where f(0) is positive. We can then write the following chain of equalities:

f(x + y) = f(x)*f(y) = f(x)*f(0) = f(0)*f(x) = f(x)

This shows that f(x) is equal to f(x) * f(0) for all x∈ℝ. Since f(0) is positive, it follows that f(x) is also positive for all x∈ℝ.

The case where f(0) is negative can be proved similarly. We can write the following chain of equalities:

f(x + y) = f(x)*f(y) = f(x)*f(0) = f(0)*f(x) = -f(x)

This shows that f(x) is equal to -f(x) * f(0) for all x∈ℝ. Since f(0) is negative, it follows that f(x) is also negative for all x∈ℝ.

Therefore, in both cases, f(x) is nonzero for all x∈ℝ.

b) f(x) > 0, for every x∈ℝ

Step-by-step explanation:

Since f(x) is nonzero for all x∈ℝ, it follows that f(x) is either positive or negative for all x∈ℝ. We know that f(0) is nonzero, so it must be either positive or negative. If f(0) is positive, then f(x) is also positive for all x∈ℝ, as we proved in part a). If f(0) is negative, then f(x) is negative for all x∈ℝ, as we also proved in part a). In either case, f(x) is positive for all x∈ℝ.

c) f(0) = 1

Step-by-step explanation:

Since f(x + y) = f(x)*f(y) for all x, y∈ℝ, it follows that f(0 + 0) = f(0)*f(0). Substituting in the values, we get:

f(0) = f(0)*f(0)

Since f(0) ≠ 0, it follows that f(0)*f(0) = 1. Therefore, f(0) = 1.

d) f(x)*f(-x) = 1, for every x∈ℝ

Step-by-step explanation:

Since f(x + y) = f(x)*f(y) for all x, y∈ℝ, it follows that f(x + (-x)) = f(x)*f(-x). Substituting in the values, we get:

f(0) = f(x)*f(-x)

Since f(0) = 1, it follows that f(x)*f(-x) = 1 for all x∈ℝ.

e) If the function f(x) = 1 has a single solution, then