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$$f(x) = \begin{cases} 3x^2+12x-1 & -1\leq x \leq 2 \\ 37-x & 2\lt x \leq 3 \end{cases}$$

This function is obviously continous at $x=2$. Also, $f'(2)$ does not exist. Before $2$, the function is increasing, because $f'$ is positive. After $2$ it is decreasing.

The question asks if the function is increasing on $[-1,2]$. The function clearly is increasing on $[-1,2)$. But, at exactly $x=2$, how should I classify it? The left hand derivative is positive and the right hand derivative is negative.

The way I see it is that it should be increasing, because on approaching $2$ from the left, the function is increasing. Also, we are not allowed to go to the right of $2$. So it must be increasing. Am I right?

Edit:

To avoid confusion with Can a function be increasing at a point?, let me clarify in that context. I don't wanna know if in general a function can be continuous at a point. I have a specific query about what happens when we restricted the period in which we are checking continuity.

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Pratyush Yadav
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    Recall the definition of increasing; we say that $f$ is increasing on $(a,b)$ if $f(x)>f(y)$ whenever $x>y$ for $x,y\in (a,b)$... – John Martin May 05 '16 at 15:40
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    Sorry, read that the wrong way round. Yes, it is increasing on $[-1,2]$. If you pick any two points $a<b$ in $[-1,2]$ you have $f(a)<f(b)$. – almagest May 05 '16 at 15:40
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    The title is "misleading" but nicely encapsulates your "point" of confusion. Related to the earlier Question in that we define increasing functions on intervals (or at least on ordered domains), and then define increasing-at-a-point to mean increasing in a suitable neighborhood of the given point (locally). – hardmath May 05 '16 at 15:54

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  1. What is a function? A subset $f$ of $A\times B$ such that for each $a\in A$ there is exactly one $b\in B$ for which $(a, b)\in f.$ A function is "its graph." The function $f$ in your first line ,and the function "$f$ restricted to the domain $[-1,2]$" are two different functions.

  2. We can say that "a function $g$ is increasing at a point $p$" (when $p\in$ dom $(g)\;$) means that there exists $r>0$ such that (i) $y\in (p-r,p)\cap$ dom $(g)\implies g(y)\leq g(p),$ and (ii) $y\in (p,p+r)\cap$ dom $(g)\implies g(y)\geq g(p)$. Then we can say that $g$ is increasing iff $g $ is inceasing at each $p$ in its domain.

DanielWainfleet
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