The discussion centers on proving that the derivative of the function g, defined by the properties g(x) = xf(x) + 1 and g(x + y) = g(x)g(y), equals g(x) for all x. The limit definition of the derivative is employed, leading to the conclusion that g'(x) = g(x) through careful manipulation of limits and the properties of the functions involved. The limit of f(x) as x approaches 0 is established as 1, which is crucial for the proof.
PREREQUISITESStudents of calculus, mathematicians, and anyone interested in understanding the relationship between derivatives and function properties, particularly in the context of functional equations.
Just use the "limit" definition of the derivative.drawar said:Homework Statement
Let f and g be two functions on ℝ such that:
1. g(x) = xf(x) + 1 for all x,
2. g(x + y) = g(x)g(y) for all x,y,
3. \mathop {\lim }\limits_{x \to 0} f(x) = 1.
Prove that: g'(x) = g(x) for all x.
Homework Equations
The Attempt at a Solution
Sorry for asking this question without showing my workings but I just don't know how to get started. Any hints would be much appreciated. TIA!
SammyS said:Just use the "limit" definition of the derivative.
drawar said:So I have to prove \mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) - g(x)}}{h} = g(x)?
SammyS said:Yes.
It works out quite nicely.