Differentiability of $f(x) = e^{-(x+2)}\sqrt{x+1}$ My students are required to study the set of differentiability of 
$$f(x) = e^{-(x+2)}\sqrt{x+1}. $$
It is of course defined in $[-1, +\infty)$ and differentiable in $(-1, +\infty)$.
They say that it is not derivable in $x=-1$ because the limit is $+\infty$, I say that it makes no sense to require for differentiability at borders, even if the limit is finite. Who is right?
They also say that $-1$ is not a local minimum because there is no interval, since $f$ is not defined for $x<-1$. I argue it is a local minimum actually. Who's right?
Thanks!
 A: You are right.
For the second one: extremas are depends on the set that $f$ exists, and because there is no derivative (it is not differentiable) at $-1$ we need to check, in general extremas can appear at critical point and not differentiable points.
For the first one, their argument is non-sense, the limit does not exists (in this case it is $+\infty$) because it is not differentiable, how can they take the derivative of a point that only the right side limit exists for? By the definition of derivative we need to have 2 sided limit to takes the derivative of this point. So again you are right

As the comments pointed out we can define derivative with one side limit, so if this is the case then if the right side limit goes to $\infty$ they are right. So it depends on if you include one side derivative or not
A: I decide to look up Bartle's Introduction to Real Analysis for the definitions.

(Definition of differentiability)


So according to the book, derivative at end points can indeed be defined as one-sided limit. Now
$$\begin{align*} 
\frac{f(x) - f(-1)}{x - (-1)} 
&= \frac{e^{-(x + 2)}\sqrt{x + 1} - e^{-(-1 + 2)}\sqrt{-1 + 1}}{x + 1} \\
&= \frac{e^{-(x + 2)}\sqrt{x + 1}}{x + 1}  \\
&= \frac{e^{-(x + 2)}}{\sqrt{x + 1}} \to +\infty \text{ as } x \to -1^+
\end{align*}$$
So I think your students can be considered correct for their first claim.

(Definition of local extremum)


Again according to the book, a function is said to have a local minimum at $c$ if there is an open interval $I$ centered at $c$ such that $f(c) \le f(x)$ for all $x$ lying in the intersection of $I$ and the domain of $f$. Now since $f$ is strictly decreasing, $f(-1) \le f(x)$ for all $x$ lying in $(-2, 0) \cap [-1, +\infty)$. So I think your students are wrong for their second claim.
