Local extrema: definitions, derivative signs, and the domain
Distinguish strict local extrema from absolute extrema; apply the first derivative test with its continuity and interior-point hypotheses.
Content updated: October 22, 2026
A local extremum compares a function value with values at nearby points; an absolute maximum or minimum compares it with values throughout the specified domain. A zero derivative identifies a point to investigate, but does not by itself establish a local extremum.
Objectives: use the definition or derivative signs, check that the point is interior to the domain, and distinguish local from absolute extrema.
Prerequisites: monotonicity (English lesson); derivatives and inequalities involving square roots.
On this page: Idea · Method · Examples · Practice · Further learning
1. A local extremum is a local property
This lesson uses strict local extrema at interior points: $f$ has a strict local maximum at $x_0$ if there is a $\delta>0$ such that $f(x)<f(x_0)$ whenever $0<|x-x_0|<\delta$; reverse the inequality for a strict local minimum. This neighborhood lies in the domain under consideration. The extremum occurs at the input $x_0$, its value is $f(x_0)$, and the corresponding point on the graph is $(x_0;f(x_0))$.
For a problem restricted to the closed interval $[a;b]$, we count only interior local extrema in $(a;b)$ under this convention. The endpoints must still be considered when finding the absolute maximum and minimum on the interval. Under a definition relative to a set, endpoints may be called one-sided local extrema; state the convention to avoid two different counts.
2. Derivative tests and their hypotheses
If $f$ has a local extremum at an interior point and is differentiable there, Fermat's theorem implies $f^{\prime}(x_0)=0$. The converse fails: for example, $x^3$ has derivative $0$ at $0$ but increases through that point. A function may also have a local extremum where it is not differentiable, as $|x|$ does at $0$.
For the first derivative test, suppose that $f$ is continuous at $x_0$ and differentiable on both sides in a punctured neighborhood. A positive derivative on the left and a negative derivative on the right imply a strict local maximum; a negative derivative followed by a positive derivative implies a strict local minimum. These conclusions follow from monotonicity on each side and continuity at the point. If the derivative has the same strict sign on both sides, there is no local extremum. Do not omit continuity when joining the two branches.
3. Five worked examples
Example 1. Completing the square
For $f(x)=x^2-x+1$, we have $f(x)=(x-1/2)^2+3/4\ge3/4$, with equality only at $x=1/2$. Thus a strict local minimum occurs at $1/2$, its value is $3/4$, and the corresponding graph point is $(1/2;3/4)$. Here the local minimum value is also the absolute minimum on $\mathbb R$.
Example 2. A local maximum from an inequality
Let $f(x)=\sqrt{x-1}+\sqrt{2-x}$, with domain $[1;2]$. Set $u=\sqrt{x-1}$ and $v=\sqrt{2-x}$; then $u,v\ge0$ and $u^2+v^2=1$. By the Cauchy–Schwarz inequality, $u+v\le\sqrt2$. Equality holds when $u=v$, or $x=3/2$. Hence $f(x)\le f(3/2)=\sqrt2$, with strict inequality if $x\ne3/2$, so there is a strict local maximum at $3/2$.
Example 3. Differentiating a quadratic function
Again consider $f(x)=x^2-x+1$. Its derivative $2x-1$ is negative before $1/2$ and positive after $1/2$. Since the function is continuous, the first derivative test confirms the strict local minimum at $(1/2;3/4)$, in agreement with Example 1.
Example 4. Differentiating a sum of square roots
For the function in Example 2, the derivative is calculated only on $(1;2)$:
It is $0$ at $3/2$, positive when $1<x<3/2$, and negative when $3/2<x<2$. Continuity therefore implies a strict local maximum at $3/2$, with value $\sqrt2$. The endpoint values are $f(1)=f(2)=1$; the endpoints are not counted as interior local extrema under this lesson's convention.
Example 5. Restriction to a closed interval
Let $f(x)=x^3-3x^2+1$ on $[0;4]$. The derivative $f^{\prime}=3x(x-2)$ is $0$ at $0,2$. Only $2$ is interior to this interval; the derivative is negative on $(0;2)$ and positive on $(2;4)$. Thus there is one interior strict local minimum, at $2$, with value $f(2)=-3$, and no interior local maximum.
On the whole of $\mathbb R$, the derivative is also positive before $0$ and negative after $0$, so $0$ is a strict local maximum point, with value $1$. Changing the domain changes which points are interior, but does not change the derivative. Finding the absolute maximum on $[0;4]$ is a separate question.
4. Practice exercises
- Does the function $x^3$ have a local extremum at $0$?
- Prove that $|x|$ has a strict local minimum at $0$ although it is not differentiable there.
- Find the local extrema of $-x^2+4x$ on $\mathbb R$.
Show answers and hints
- No. The derivative $3x^2$ is positive on both sides, so the function still increases through $0$.
- Since $|x|>0=|0|$ for every $x\ne0$, the definition implies a strict local minimum at $0$.
- The derivative $-2x+4$ changes from positive to negative at $2$; there is a strict local maximum at $2$, with value $4$.
5. Common mistakes
- Calling a zero of the derivative a local extremum without checking signs.
- Confusing the input, the value, and the point on the graph associated with a local extremum.
- Counting an endpoint as an interior point without stating a convention.
- Using the sign-change test for a function that is discontinuous at the joining point.
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