Does $f$ have an Extremum at $x = c$ if $c$ is Not in the Domain of $f'$?

Question

I have a question that is probably silly to many.

I was given the function $f(x)=x^2-|2x-1|$ and was requested to find its extrema. Notice that $f'(x)=2x-\frac{2(2x-1)}{|2x-1|}$, so that $f$ is not differentiable in $x=0.5$ and this therefore constitutes a critical point.

To find the extrema I solved $f'(x)=0$, which gives three results: {$1, -1, 0.5$}. But $0.5$ must be excluded, since it's not on the domain.

My question is the following. Since $0.5$ is on the domain of $f$, and is a solution for $f'(x)=0$, is $x=0.5$ an extremum? My first instinct was to say no; that this point would only constitute a critical point, but not an extreum, since $x=0.5$ is not really a root of the derivative, because it's not part of its domain. But then I saw the graph of $f$ and $x=0.5$ was indeed an extremum! I'm very confused about this, could anyone clarify?

Answer 1

What you write is not consistent. You write that $f$ is not differentiable at $\frac12$. I agree. But that's the same thing as asserting the $\frac12$ doesn't belong to the domeain of $f'$. However, you write later that $\frac12$ is a solution of the equation $f'(x)=0$. There is a contradiction here.

In fact, $f$ is not differentiable at $\frac12$ and therefore $\frac12$ is not a solution of the equation $f'(x)=0$. However, $f$ may very well have an extrema there. Take, for instance, the function $g(x)=\lvert x\rvert$. It attains its minimum at $0$, a point at which $g$ is not differentiable.

Answer 2

An extremum is simply defined as the largest or smallest value of the function in the interval. For differentiable functions, this coincides with points where $f'(x)=0$ since, as I'm sure you know, those are turning points. But just because we can infer an extremum from $f'(x)=0$, that doesn't mean that it's a necessary condition since plenty of functions aren't differentiable. So yes, we can have an extremum without $f'(x)=0$.

Easy example: $f(x)=\begin{cases}x+1&x\leq0\\0&x>0\end{cases}$

It's plain to see there's a local maximum of $1$ at $x=0$. But the function isn't even continuous, let alone differentiable there so we have an extremum without $f'(x)=0$.

Answer 3

The definition of "extremum" has nothing to do with derivatives. A point $x=x_0$ is a maximum for $f$ if $f(x_0)\ge f(x)$ for all $x$, period. Similarly for minima, and now an extremum is a point that is either a minimum or a maximum.

Let's consider a simpler example: $f(x)=|x|$. It's clear that $f(0)\le f(x)$ for all $x$, hence $x=0$ is a minimum for $f$; the fact that $f'(0)$ does not exist is irrelevant.

Let's assume below that $x_0$ lies in the interior of the domain of $f$, meaning that there exists $r>0$ such that $f(x)$ is defined for every $x$ with $x_0-r< x<x_0+r$. The theorem is this:

Suppose $f$ has an extremum at $x_0$. If $f$ is differentiable at $x_0$ then $f'(x_0)=0$.

Note the "If"! The result is the same as this:

If $f$ has an extremum at $x_0$ then either $f'(x_0)$ does not exist or $f'(x_0)=0$.

At the extrema in the examples above the derivative does not exist; hence "either $f'(x_0)$ does not exist or $f'(x_0)=0$" is true.