Existence of the derivative of a function at an end point For a function with the rule $f(x)=x^3e^{-2x+1}$, where $x\ge 0$,
I thought that the derivative does not exist at $x=0$.
My friend disagreed with me and told me that the derivative exists at this point.
Can anyone enlighten me on this?
Thanks
 A: For $x \ne 0$ we have
$$\frac{f(x)-f(0)}{x-0}= \frac{x^2}{e^{2x-1}}.$$
Hence $ \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}$ exists and $=0.$
A: Differentiability is usually defined in the interior of the domain, and $\operatorname{int}([0,\infty))=(0,\infty)$. So strictly speaking, $f$ is not differentiable at $x=0$.
Differentiability on an open interval is defined something like this: We call $f:\mathbb{R}\to \mathbb{R}$ differentiable on $(a,b)$ if $\forall x \in (a, b)$, we have that $\exists f'(x)$.
And you can extend this definition to closed, half-closed, etc. sets as well. For example, differentiability on $[a, b)$ is usually defined as follows: $f:\mathbb{R}\to \mathbb{R}$ differentiable on $[a,b)$ if $\forall x \in (a, b)$, we have that $\exists f'(x)$ and $\exists \lim\limits_{x \to a + 0} \frac{f(x)-f(a)}{x-a}$.
So it's up to your definition if you call it a differentiable function on $[0, \infty)$ or not. But it's not differentiable at $x=0$ in the usual sense, just from the right side.
A: If you say that $f$ is defined only in $[0, +\infty[$, then $f$ is right-differentiable in $0$ because the one-sided limit
$$\lim_{x \to 0^{\color{red}+}} \frac {f(x) - f(0)} {x - 0}$$
exists as a real number.
A: That exponential stuff is a red herring here.  Let $g$ be any bounded function defined on a neighbourhood of $0$.  Then obviously $f(x):=x^2\cdot g(x)$ is differentiable in $0$ and $f'(0)=0$ as
$$\frac{f(x)-f(0)}{x-0}=x\cdot g(x).$$
