# If a first derivative doesn't exist at a certain point, is it not a critical point?

Say $$f'(x)$$ of a function was $$(x+2) \over (x+3)$$. If $$x = -2$$ then $$f'(x)$$ = 0, which means that at this point, there would be a local min or max. But what if $$x = -3$$? It doesn't exist for $$f'(x)$$, so do we just ignore it? State it DNE at $$x = -3$$ so it is not a critical point?

• Your function is not defined for $x=-3$. – hamam_Abdallah Nov 29 '18 at 20:11
• The derivative is $f'(x)=1/(x+3)^2$. It is not zero at $x=-2$. – Klaas van Aarsen Nov 29 '18 at 20:18
• The proper term for $x=-3$ is singularity rather than critical point. – Klaas van Aarsen Nov 29 '18 at 20:19
• Thanks! also one more thing, when the second derivative is 0, does the concavity always change at that point? – ming Nov 29 '18 at 20:28
• No, concavity does not necessarily change at the point $f’’(x) = 0$. Take $f(x) = x^4$ as an example. Concavity changes at the point where $f’’(x)$ changes signs. – KM101 Nov 29 '18 at 20:42

As is often the case in mathematics, it comes down to starting with a good, solid definition. Here are a few reasonable places to start:

From Wikipedia:

In mathematics, a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0. Some authors also classify as critical points any limit points where ... the derivative is not defined. [emphasis mine]

From MathWorld:

A function $$y=f(x)$$ has critical points at all points $$x_0$$ where $$f'(x_0)=0$$ or $$f(x)$$ is not differentiable. A function has critical points where the gradient $$\nabla f = \mathbf{b}$$ or $$\partial f /\partial x$$ or the partial derivative $$\partial f / \partial y$$ is not defined.

From Thomas' Calculus:

An interior point of the of the domain of a function $$f$$ where $$f'$$ is zero or undefined is a critical point of $$f$$. [bold emphasis mine]

Note that both Wikipedia and Thomas' Calculus define critical points of a function $$f$$ to be points in the domain of $$f$$ (Thomas goes a step further and only considers points in the interior of the domain). The MathWorld definition does not make this restriction, nor is its definition of differentiability sufficiently precise to require any restriction.

We should also consider the goal of defining a critical point. In elementary calculus classes, the goal is usually to move on to optimization problems. In optimization problems, we look for extrema (a) at places where the derivative is zero (Fermat's theorem), (b) at boundary points in the domain, and (c) at domain points where the derivative fails to exist. Thus we might want to define critical points so that we see these three kinds of behaviour. Hence I would propose that critical points are properly defined as follows (for the purposes of a calculus class):

Definition: A critical point of a function $$f$$ is a point $$x_0$$ in the domain of $$f$$ such that either $$f'(x_0) = 0$$ or $$f'(x_0)$$ is undefined.

Note that the derivative is usually considered to be undefined on the boundary of a set, so this definition picks up on all three of the kinds of points that we are interested in without causing problems.

Finally, applying this definition to your question, $$x_0 = 3$$ is not a critical point of $$f(x) = \frac{x-2}{x-3},$$ since $$x_0 = 3$$ is not in the domain of $$f$$, at least not as the domain is usually understood (we could, I suppose, work over the extended real numbers).

• Good presentation of the case. Thanks. – NoChance Nov 30 '18 at 19:41
• Great explanation! Thank you – Achuan Chen Jan 7 '20 at 15:27

I think the easiest explanation at a basic calculus level is to simply say that no, it is not a critical point. Hence if we want to do problems regarding optimization for a function which does not have its derivative (or the function itself) defined everywhere, then you would want to consider the "shape" of that function when deducing things, and not always just writing $$f'(x)=0$$. For example, if you want to minimise $$1/x$$, setting its derivative equal to $$0$$ you get $$1=0$$, which leads you to the conclusion that there are no critical points. Indeed, if you look at the graph, there aren't any "hills" or "valleys". Yet it still makes sense to say that its minimum value is arbitrarily large, or sometimes $$-\infty$$, in many contexts.

The derivative of $$|x|$$ does not exist at $$0$$. Still, that is the only local extrema, so you better not throw it away...

Quoting Wikipedia :

In mathematics, a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is $$0$$. Some authors also classify as critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.

The above definition can be found here.

Thus, it all really comes down to what you want to define as a critical point. The most standard definition (used in dynamical systems for example) is in the case of the derivative being equal to zero. If that's the case, no, it's not a critical point.

On the other hand, if you want to include to the definition of a critical point any points where the derivative is not defined, then yes, $$x=-3$$ is a critical point.

• I think you really ought to reference the source of the quote. – user296602 Nov 29 '18 at 20:23
• @T.Bongers Added, thanks for the heads up. That's why I added the italic. – Rebellos Nov 29 '18 at 20:25
• Sure, no problem. But per Wikipedia's licensing, it's best to include a hyperlink directly to the source. – user296602 Nov 29 '18 at 20:26
• Thanks! I think my course AKNOWLEDGES that $x = 3$ is an important point, but we wouldn't say it is a CRITICAL POINT because it is not a local extrema. Is that pretty standard? – ming Nov 29 '18 at 20:28
• @ming Well, the terms used in your sentence aren't that rigorous so I wouldn't comment it as standard. The most used definition is that if $f'(x_0) = 0$ then $x_0$ is a critical point. – Rebellos Nov 29 '18 at 20:30