# Functions-4

### From Homeworkwiki

**Let f : R → R be such that f(x) = 2 ^{x}. Determine (a) Range of f (b) {x : f(x) = 1}**

**(c) whether f(x+y) = f(x).f(y) holds.**

**Solution:** (a) Since 2^{x} is positive for every x ∈ R.

So, f(x) = 2^{x} is a positive real number for every x ∈ R.

Moreover, for every positive real number x, there exist log_{2} x ∈ R such that f(log_{2} x) = 2^{log2x} = x

Hence, we conclude that the range of f is the set of all positive real numbers.

(b) Since, f(x) = 1 => 2^{x} = 1 => 2^{x} = 2^{0} => x = 0

Therefore, {x : f(x) = 1} = {0}.

(c) Since, f(x + y) = 2^{x + y} = 2^{x} 2^{y} = f(x) f(y)

Therefore, f(x + y) = f(x) f(y) holds for all x, y ∈ R.