A function $f: \mathbb{R} \to \mathbb{R}$ has property $f(x+y) = f(x) \cdot e^{f(y)-1}$, for every $x, y \in \mathbb{R}$ then positive value of $f(4)$ is:
Step-by-Step Solution
Key Concept: From the functional equation f(x+y) = f(x)·e^(f(y)-1), setting y=0 gives f(x) = f(x)·e^(f(0)-1), which implies f(0)=1. Then substituting x=y yields f(2y) = f(y)·e^(f(y)-1), and analyzing the helper function g(t) = t - e^(t-1) shows that f(y)=1 for all y is the only solution where f remains consistent.
Note that $f(0) = 1$ and putting $x=0$ in the functional equation gives $f(y) = f(y)^{-1}$. Consider the helper function $g(t) = t - e^{t-1}$. Here $g'(t) 1$ and $g'(t) > 0$ if $t < 1$, and $g(1) = 0$. So $g(t)$ has only one root at $t = 1$.
Correct Answer: 1