Same derivative of two functions in a point

Question

First of all: sorry for the vague title. I am thinking about the following: Let $f,g:\mathbb R \to \mathbb R$ be $C^1$ functions and there exists a point $x_0 \in \mathbb R$ such that $f(x_0) = g(x_0)$ and $f'(x_0)=g'(x_0)$. Is this enough to say that there exists an $\varepsilon >0$ such that $f(x) = g(x)$ for all $x \in (x_0-\varepsilon, x_0+\varepsilon$? Or do I need stronger conditions?

Answer 1

Consider the taylor expansions of $f$ and $g$ around $x_0$ in an $\epsilon$ neighbourhood. The first two terms are the same but unless the other terms are same as well, they are not equal.

Answer 2

A counterexample has been given, but the reverse is true: If $f(x)=g(x)$ for all $x\in(x_{0}-\epsilon,x_{0}+\epsilon)$, then \begin{align*} f'(x_{0})&=\lim_{h\rightarrow 0,~|h|<\epsilon}\dfrac{f(x_{0}+h)-f(x_{0})}{h}\\ &=\lim_{h\rightarrow 0,~|h|<\epsilon}\dfrac{g(x_{0}+h)-g(x_{0})}{h}\\ &=g'(x_{0}). \end{align*}

Answer 3

A counter example you can look at is $g(x)=x,\ f(x)= 2^{-1}x^2$, $x_0=\ ?$.

Answer 4

Consider $h(x) = f(x) - g(x).$ The conditions of this question are that $h(x_0) = h'(x_0) = 0.$ Are those conditions sufficient to ensure that $h(x) = 0$ everywhere in a neighborhood around $x_0$?