# Is it correct to say all extrema happen at critical points but not all critical points are extrema?

Is it correct to say all extrema happen at critical points but not all critical points are extrema?

This question is for single variable calculus. But it would be great if someone could also provide insight to how well this statement generalizes to more complex functions (multi-variable functions, vector-valued functions, etc)

• Extrema need not be critical points. They can also be the "end-points" in a given domain. This is what is called "absolute extrema". – Airdish Mar 31 '16 at 10:56
• All interior extrema are critical points. Of course, critical points need not be extrema. – Sangchul Lee Mar 31 '16 at 10:58

Every local extremum in the interior of the domain of a differentiable function is neccesarily a critical point, i.e. $f'(x)=0$ is a necessary condition for $x$ to be a local extremum. There are critical points which are not local extrema.

Note that I'm stressing local extremum in the interior here. To account for global extrema, one needs to consider the behavior on the boundary of the domain as well.

Consider the function $$f:[-1,1]\rightarrow \mathbb R, x\mapsto x^2.$$ $x_0=0$ is a critical point and indeed a local extremum (a minimum). For the global extrema note that $x^2 \leq 1$ for $x\in [-1,1]$ so that $f(1)=f(-1)=1$ and $x_1=1, x_2=-1$ are points where $f$ attains a (global) maximum, but they are not critical points.

Similarly, for $$f:[-1,1]\rightarrow \mathbb R, x\mapsto x^3.$$ $x_0=0$ is a critical point, but there's no extremum. Maximum and Minimum are attained at $x_1=1$ and $x_2=-1$, neither of which are critical points.

• When you say boundary points. You just mean boundary to domain (builtin or contrived, right?) What I mean is $2$ would be an inherent boundary point to $x/(x-2)$ and contrived ones would be $[-1,1]$. Sorry if this is not correct terminology. Or am I just thinking too hard? And the inherent ones are already covered by the $f'(c)=0$ or DNE clause? – AlanSTACK Apr 1 '16 at 1:13
• @Alan: I mean boundary points of the domain where the function is defined. – Roland Apr 1 '16 at 5:23

If you assume $f$ is a continuous, real-valued function on a closed, bounded interval $[a, b]$, then:

• $f$ has at least one absolute maximum and at least one absolute minimum. (The Extreme Value Theorem.)

• If $f$ has a local (relative) extremum at an interior point $x_{0}$, then either $f'(x_{0}) = 0$ or $f'(x_{0})$ does not exist. (I.e., $x_{0}$ is a critical point of $f$, modulo your definition of "critical point".)

• If your definition of "critical point" includes endpoints of $[a, b]$, then "yes", every local extremum of $f$ is a critical point of $f$. (This convention is reasonable, since strictly speaking $f'$ does not exist at an endpoint of an interval. On the other hand, this convention is not universal.)

For the other direction:

• An interior critical point can fail to be a local extremum. Examples (with $x_{0} = 0$) include $f(x) = x^{3}$ and $$g(x) = \begin{cases} x & x < 0, \\ 2x & x \geq 0, \end{cases}$$ both viewed as functions on some interval containing $0$ in the interior. (In the first example, $f'(0) = 0$; in the second, $g'(0)$ does not exist.)

• An endpoint can fail to be a local extremum, e.g. $$f(x) = \begin{cases} x^{2} \sin(1/x) & 0 < x, \\ 0 & x = 0. \end{cases}$$

There are analogous results for real-valued functions of more than one variable, but they're a little more work to state, because not every "connected subset" of the plane (say) is a formal analog of a closed, bounded interval. Loosely, though, a continuous function $f$ on a closed, bounded subset $D$ of $\mathbf{R}^{n}$ has at least one absolute maximum and at least one absolute minimum in $D$, and these must occur either at (i) an interior point $x_{0}$ of $D$ where the total derivative $Df(x_{0})$ is zero; (ii) an interior point $x_{0}$ where $Df(x_{0})$ does not exist; or (iii) a boundary point of $D$.

There is no analog for vector-valued functions because the concepts of "maximum" and "minimum" make no sense. (If $n > 1$, the set $\mathbf{R}^{n}$ is not ordered in any way compatible with vector addition and scalar multiplication.)

• For the left endpoint $a$, the differential quotient $\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}$ is a limit which may (or may not) exist, and should be regarded as a one-sided limit (i.e. only for $h>0$). If $f'(a)$ exists or not is dependent on the existence of this one-sided limit. – Roland Mar 31 '16 at 11:38
• @Roland: Agreed, but in my experience that criterion of differentiability at an endpoint is (also) a convention, and (with some justification) not a universal one. (If it matters, I'm not advocating either convention of endpoint differentiability, just noting that as stated, the OP's question can only be answered with qualifications.) – Andrew D. Hwang Mar 31 '16 at 11:55
• yet you write 'strictly speaking $f′$ does not exist at an endpoint of an interval', which is wrong for functions which are differentiable in a larger open domain which includes the interval $[a,b]$. – Roland Mar 31 '16 at 12:02
• Strictly speaking, "$f$ is differentiable at $x_{0}$" involves a two-sided limit at $x_{0}$. If you extend the definition at an endpoint by requiring only a one-sided limit, that's fine, but you're no longer invoking the usual definition of differentiability from one-variable calculus. (If you're saying the extended definition is universal, I have no mathematical grounds to argue otherwise. I'm merely saying that this extension is not tacit in elementary calculus. For instance, $f(x) = |x|$ is not differentiable at $0$, but is differentiable on $[0, \infty)$ by your definition.) – Andrew D. Hwang Mar 31 '16 at 12:36
• @Roland It is explicitly stated in some real analysis textbooks (ex. Apostol) that for a derivative $f'(c)$ to exist, $f$ must be defined in an open interval containing $c$. Other authors, like Rudin, do not explicitly require this yet mention the issue of differentiability at endpoints in passing. The motivation for this requirement is far beyond the scope of introductory calculus. See the answer by Willie Wong here: math.stackexchange.com/questions/126176/… – MathematicsStudent1122 Apr 2 '16 at 5:07

Yes.

You can analyse this multi-variable example $z = x^2 - y^2$.

This has a critical point which is not an extrema. These points called saddle points.

• What's your function? What's your domain? – Roland Mar 31 '16 at 11:40
• @Roland $f: \mathbb R^2 \rightarrow \mathbb R$ with $z = f(x,y) = x^2 - y^2$. – crbah Mar 31 '16 at 11:44
• What if the domain is $\{(x,y): x^2 +y^2 \leq 1\}$? Is the answer still yes? – Roland Mar 31 '16 at 11:49
• @Roland Yes. You can see its graph on en.wikipedia.org/wiki/Saddle_point – crbah Mar 31 '16 at 11:53
• My comment for the restricted domain is not about the critical point which is not an extremum. It's about the extremal points which are not critical points ($(x,y)=(1,0)$ and $(x,y)=(0,1)$). – Roland Mar 31 '16 at 12:00