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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)

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3 Answers

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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.

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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. $$ f(x) = \begin{cases} x^{2} \sin(1/x) & 0 < x, \\ 0 & x = 0. \end{cases} $$

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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.)

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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.

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