Just use the "limit" definition of the derivative.drawar said:Homework Statement
Let f and g be two functions on ℝ such that:
1. [itex]g(x) = xf(x) + 1[/itex] for all x,
2. [itex]g(x + y) = g(x)g(y)[/itex] for all x,y,
3. [itex]\mathop {\lim }\limits_{x \to 0} f(x) = 1[/itex].
Prove that: [itex]g'(x) = g(x)[/itex] 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 [itex]\mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) - g(x)}}{h} = g(x)[/itex]?
SammyS said:Yes.
It works out quite nicely.
The derivative g'(x) is the instantaneous rate of change of a function g(x) at a specific point x. It represents the slope of the tangent line to the graph of g(x) at that point.
To prove that g'(x)=g(x), you can use the definition of the derivative to show that g'(x) is equal to the limit of (g(x+h)-g(x))/h as h approaches 0. Then, you can use algebraic manipulation and substitution to show that this limit is equal to g(x). This will prove that g'(x)=g(x).
Sure, let's take the function g(x)=x^2. To prove that g'(x)=g(x), we need to show that the limit of [(x+h)^2-x^2]/h as h approaches 0 is equal to x^2. By expanding the numerator and simplifying, we get the limit of 2x+h as h approaches 0, which is equal to 2x. Therefore, g'(x)=2x=g(x), proving the equality.
Some common mistakes when proving g'(x)=g(x) include incorrect algebraic manipulation, not using the correct definition of the derivative, and not considering the limit as h approaches 0.
Proving g'(x)=g(x) is important because it shows that a function is differentiable, meaning it has a well-defined derivative at all points in its domain. This allows us to use the derivative to solve various problems in calculus and other fields of science and engineering.