Prove to me that this answer is correct

Question

I'm doubting myself and my teacher, so I need some external verification.

So, the graph on the left is given to me, and I'm told to derive it. The answer key is shown on the right. What I do not understand is why $x = -5$, $-1$ and $3$ are not inclusive.

I understand that the limits at $x = -5$, $-1$ and $3$ do not exist, obviously. However, they are still derivable at those points, correct? Because in the original graph, they are clearly marked inclusive. It even says the domain is $[-5, \infty)$, so that completely means $-5$ is inclusive for $f(x)$. (That infinity one should really be a right parenthesis.)

So anyway, is my teacher wrong, or am I wrong? Can anyone actually prove to me in the most clear, concise manner possible that $f'(x)$ at the points $-5$, $-1$ and $3$ are non-inclusive?

Thank you for your time and help in advance. It is much appreciated.

Answer 1

The definition of the derivative of $f$ at a point $x = c$ is $$ \lim_{x\to c}\frac{f(x) - f(c)}{x-c} $$ One can discuss whether this is defined at the boundary of the domain of the function (it's conventional to say that it isn't, for convenience). However, when the function has a discontinuity the limit definitely isn't defined.

If $c$ is a point of discontinuity, then there are values for $x$ which are arbitrarily close to $c$ where the numerator of the above fraction is nowhere close to $0$. Since the denominator is close to $0$ when $x$ is close to $c$, that means that the limit doesn't exist. In this case we can't even say that the limit "is" $\infty$, because there are also values for $x$ close to $c$ for which the numerator is close enough to $0$ to give a finite value.

Alternatively, geometrically, the derivative of $f$ at $c$ is the slope of a linear function $g(x) = ax + b$ such that for values of $x$ which are close to $c$ (including being equal to $c$ this time, since there are no denominators in this interpretation), the value of $f(x)$ is close to the value $g(x)$ (for a certain, strictly defined notion of "close"). We can see that for the points of discontinuity you have here, while there is such a line for each side separately, there is no such line that comes close to the graph of $f(x)$ on both sides at the same time. Therefore the function doesn't have a derivative at those points.

Answer 2

Note that we can define derivative at any point $-5+\epsilon$ and $-1-\epsilon$ $\forall \epsilon$ with $0<\epsilon<1$ but at $x=-1$ and $x=-5$ derivative are not defined since the limit of incremental ratio is not defined.