Does $f(x) = |x|$ have a local minimum?

Question

I would like to know whether $f(x) = |x|$ has a local minimum at $x = 0$. For a local minimum, the first derivative of the function has to be zero. However, in the case of $|x|$, the function is not differentiable at $x = 0$. Will the point (0,0) still be considered as a local minimum?

Answer 1

You do not need derivatives here because you have $f(x)=|x|>0$ for every $x \in \mathbb R \setminus \{0\}$ and $f(0)=0$ so $0$ is really local and global minimum.

For example, the function $f(x)=\begin{cases} \frac {1}{x^2} &\text{ if } x \neq 0\\0 &\text{ if } x =0.\end{cases}$ has local minimum at $0$ and not only that it is not differentiable at $0$ but it is not continuous at $0$.

Answer 2

Yes it is a local minimum, end even a global minimum, as $$\forall x\in\mathbb{R},\, f(x)\geq f(0)$$ The function does not need to be differentiable to have local minimum. What is true is that if it's differentiable and if $x$ is a local minimum, then $f'(x)=0$.

Note that the converse is false. Counter-example: $f(x)=x^3$. We have $f'(0)=0$ but $0$ is not a local minimum.

Answer 3

You have it the wrong way. A function doesn't have to be differentiable (at a point) to attain a minimum or maximum there. Look up the definition of 'minimum' and 'maximum' in your book or course notes; it is (only) about the neighboring function values.

However: if a function is differentiable, then you have a minimum or maximum when the derivative changes sign there (careful: a zero derivative isn't sufficient; e.g. $f(x)=x^3$).

In your example, $|x|$ attains a local (and even a global) minimum at $x=0$, although the derivative doesn't exist there.

Answer 4

You have to read the fine print. "For a local minimum the first derivative has to be zero" isn't quite true. What's true is this:

If $f$ has a local minimum at $x$ and if $f$ is differentiable at $x$ then $f'(x)=0$.

The function $|x|$ obviously has a minimum at $x=0$; that does not contradict the above precisely because it's not differentiable there.

Answer 5

Yes, 0 is a local minimum of that function, since there exists a $k > 0$ such that for all $\epsilon$ with $k > \epsilon > 0$ there is $f(x-\epsilon) > f(x)$ and $f(x+\epsilon) > f(x)$.

The function is indeed not differentiable at $x = 0$, however, this is not mandatory for an extremum.

Answer 6

You have it backwards. If $f'(x_0)= 0$ and $f''(x_0) \gt 0$ then there is a local min at $x = x_0$. Because $f(x) = |x|$ is not differentiable at $x=0$, then the derivative test does not apply there. Still it is pretty clear that if $x \in (-\delta, \delta)$ then $f(0) \le f(x)$. Hence there is a local minimum at $x = 0$.

There can even be a local minimum at $x = x_0$ where the function is differentiable at $x = x_0$ and $f'(x_0) \ne 0$. Consider the function $f(x) = x$ defined on the interval $[0,3]$. For this function, there is a global minimum at $x = 0$ even though $f'(0) = 1$.

Answer 7

First, it is not "a local minima" because "minima" is plural. You want "a local minimum", the singular. That's trivial but it makes me wonder if you really understand what these words mean. The definitions of "maximum" or "minimum" have nothing to do with a the derivative being $0$. A particular $x= a$ gives a "local minimum" for $f$ is if $f(a)\le f(x) \ \forall x \in \mathcal{B}_\delta(a)$ where $\delta \rightarrow 0^+$.

(In mathematics, which has so many "formulas", there is a grave danger of confusing "learning mathematics" with "memorizing formulas". Do not fall into that trap! Understanding is better than memorizing.)