Differentiability of $f(x) = e^{-(x+2)}\sqrt{x+1}$

Question

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!

The correct term is differentiability. A function can be differentiable, not derivable. — , Dec 13 '17 at 18:47
It depends on the definition used, some books/authors define derivative only on open intervals, while some books/authors also define derivative at the end point using one-side limit. — Alex Vong, Dec 13 '17 at 18:50
You may want to read this: https://en.wikipedia.org/wiki/Semi-differentiability — , Dec 13 '17 at 18:51
The definition of minimum does not depend on the derivative, there is a minimum at $x=-1$. — Wintermute, Dec 13 '17 at 18:51
yes, both. You can look up the standard definition in any precalc book. — Wintermute, Dec 13 '17 at 18:53
@Maffred You can easily check the minima at $x=-1$. The minimum value of the function possible is $0$ i.e., at $x=-1$. Otherwise, for very large values of $x$, the function tends to $0$ but is never exact. So, $x=-1$ will be the local and the global minima — Your IDE, Dec 13 '17 at 18:56

ℋolo · Answer 1 · 2017-12-13T19:02:01.793

1

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

edited Dec 13 '17 at 19:02

answered Dec 13 '17 at 18:55

ℋolo

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2

Actually it is a matter of convention whether you allow a derivative to be defined at the edge of a domain. It's perfectly legitimate to consider the one-sided limit of $f((x+\epsilon)-f(x))/\epsilon$, although you may not want to. – gj255 Dec 13 '17 at 18:58
@gj255 I see, I'll edit the answer to be better. Thanks – ℋolo Dec 13 '17 at 18:59

Alex Vong · Answer 2 · 2017-12-13T20:00:28.870

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.

Differentiability of $f(x) = e^{-(x+2)}\sqrt{x+1}$

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