What Is A Critical Point?
A critical point of a function $f$ is a point $x = c$ inside the domain of $f$ where either the derivative is zero or the derivative fails to exist:
$$f'(c) = 0 \qquad \text{or} \qquad f'(c) \ \text{does not exist}.$$
Two conditions, one list. The first case, $f'(c) = 0$, is a place where the graph has a horizontal tangent line, the slope has flattened out. The second case is a place where the graph has a corner, a cusp, or a vertical tangent, so no single slope can be assigned. A point where $f'(c) = 0$ is also called a stationary point, since a particle moving along the curve would have zero instantaneous rate of change there.
Two details separate a correct definition from a sloppy one:
The point must be in the domain. If $f$ is not even defined at $c$, then $c$ is not a critical point. For $f(x) = \frac{1}{x}$, the value $x = 0$ is not a critical point, because the function has no value there to begin with.
Critical points are interior points. The endpoints of a closed interval are handled separately. They can hold the highest or lowest value of a function, but they are not called critical points.
The derivative is the whole engine here, so the definition of a critical point is really a statement about slope: the slope is either flat or ill-defined.
Why Do Critical Points Matter?
Critical points matter because they are the only interior places a function can reach a local maximum or a local minimum. That claim is a theorem, not a slogan.
Fermat's theorem (interior extrema). If $f$ has a local maximum or local minimum at an interior point $c$, and $f'(c)$ exists, then $f'(c) = 0$.
Turn that around. An interior local extremum forces the derivative to be either zero or nonexistent at that point. So instead of checking every one of the infinitely many points in a domain, you only have to check the critical points. That is what makes them the engine of optimization: the shortlist of suspects for "where is this biggest or smallest."
The geometric reading is the reason to believe it. At the top of a smooth hill, or the bottom of a smooth valley, the tangent line is level, so the slope is $0$. If the graph had a nonzero slope at a peak, it would still be climbing, which means a point just to the right would be higher, contradicting the idea that you were already at the top. A flat tangent is the fingerprint of a smooth turning point.
How Do You Find Critical Points?
Finding critical points is a three-step routine you can apply to almost any function.
Differentiate. Compute $f'(x)$.
Solve $f'(x) = 0$. These are the stationary points, the horizontal-tangent candidates.
Find where $f'(x)$ is undefined but $f$ is defined. These are the corner, cusp, and vertical-tangent candidates that step 2 will never catch on its own.
The single most common way to lose marks is to run step 2 and quit. Rational powers, roots, and absolute values all produce critical points of the second kind, and they are invisible to "set the derivative to zero."
Example 1: Find and classify the critical points of $f(x) = x^3 - 3x$.
Differentiate:
$$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1).$$
Set $f'(x) = 0$:
$$3(x - 1)(x + 1) = 0 \implies x = 1 \ \text{or} \ x = -1.$$
The derivative is a polynomial, so it is defined everywhere; there are no extra undefined-derivative points. The critical points are $x = -1$ and $x = 1$. Classifying them (with the second-derivative test in the next section) gives a local maximum at $(-1, 2)$ and a local minimum at $(1, -2)$.
Final answer: critical points at $x = -1$ and $x = 1$.
Example 2: Find the critical points of $f(x) = x^{2/3}$.
Differentiate using the power rule:
$$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3,\sqrt[3]{x}}.$$
Now solve $f'(x) = 0$. The numerator is the constant $2$, which is never zero, so there is no solution to $f'(x) = 0$, and step 2 finds nothing. Step 3 saves the problem: $f'(x)$ is undefined at $x = 0$ (division by $\sqrt[3]{0} = 0$), while $f(0) = 0$ is perfectly well defined. So $x = 0$ is in the domain but has no derivative.
$$x = 0 \ \text{is a critical point, because } f'(0) \ \text{does not exist}.$$
Geometrically the graph has a sharp cusp at the origin with two branches that rise almost vertically. For $x < 0$ the slope is negative, and for $x > 0$ the slope is positive, so the function falls to the origin and then climbs. That makes $x = 0$ a local (and global) minimum, even though the derivative there does not exist.
Final answer: one critical point, at $x = 0$.
How Do You Use The First And Second Derivative Tests?
Once you have the critical points, two standard tests classify each one as a local maximum, a local minimum, or neither.
The first derivative test reads the sign of $f'$ on each side of a critical point $c$:
If $f'$ changes from positive to negative at $c$, the function rises then falls, so $c$ is a local maximum.
If $f'$ changes from negative to positive at $c$, the function falls then rises, so $c$ is a local minimum.
If $f'$ keeps the same sign on both sides, $c$ is neither.
This test connects directly to where a function is climbing or descending; a fuller treatment of that idea lives at increasing and decreasing intervals.
The second derivative test uses concavity instead of sign changes, and it applies only when $f'(c) = 0$ and $f''(c)$ exists:
If $f''(c) > 0$, the graph is concave up (a valley), so $c$ is a local minimum.
If $f''(c) < 0$, the graph is concave down (a hill), so $c$ is a local maximum.
If $f''(c) = 0$, the test is inconclusive, and you fall back to the first derivative test.
Table: How the two derivative tests classify a critical point $c$.
Test | What you check | Local maximum | Local minimum | Inconclusive / neither |
|---|---|---|---|---|
First derivative test | Sign of $f'$ around $c$ | $+$ then $-$ | $-$ then $+$ | No sign change |
Second derivative test | $f''(c)$ (needs $f'(c)=0$) | $f''(c) < 0$ | $f''(c) > 0$ | $f''(c) = 0$ |
Example 1 (continued): Classify $x = -1$ and $x = 1$ for $f(x) = x^3 - 3x$.
The second derivative is:
$$f''(x) = 6x.$$
Evaluate at each critical point:
$$f''(-1) = -6 < 0 \implies \text{local maximum at } x = -1.$$
$$f''(1) = 6 > 0 \implies \text{local minimum at } x = 1.$$
The function values are $f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$. So the local maximum is $(-1, 2)$ and the local minimum is $(1, -2)$, matching the graph on the left above.
Example 3: Show that $x = 0$ is a critical point of $f(x) = x^3$ but not an extremum.
Differentiate:
$$f'(x) = 3x^2.$$
Solve $f'(x) = 0$: the only solution is $x = 0$, so $x = 0$ is a critical point. Now test it. The second derivative is $f''(x) = 6x$, and $f''(0) = 0$, so the second derivative test is inconclusive. Use the first derivative test instead:
$$f'(x) = 3x^2 \ge 0 \quad \text{for every } x,$$
so $f'$ is positive on both sides of $0$ and never changes sign. The function is increasing on both sides, so $x = 0$ is neither a maximum nor a minimum. It is an inflection point, a place where concavity flips from concave down to concave up.
Final answer: $x = 0$ is a critical point of $x^3$, but it is not an extremum.
Example 3 is the single most important cautionary tale of the topic: a critical point is a candidate for an extremum, never a guarantee of one.
Why Does The Critical Point Rule Work?
The rule works because of what a derivative measures: the instantaneous slope of the tangent line. Read geometrically, the two cases in the definition are the only two ways a smooth turning point can happen.
Flat tangent ($f'(c) = 0$). At a peak or a valley of a smooth curve the tangent line is horizontal. Any nonzero slope would mean the curve is still going up or still going down, so the point could not be a turning point yet.
No tangent slope ($f'(c)$ undefined). At a corner, a cusp, or a vertical tangent, the curve changes direction so abruptly that no single slope fits. A V-shaped bottom like $y = |x|$ turns around at a point with no derivative, yet it is clearly a minimum.
Fermat's theorem packages both cases into one clean statement: an interior extremum can only sit where the slope is zero or missing. The theorem does not promise the reverse, though: a flat tangent can also appear at an inflection point like $x^3$, where the curve pauses its climb for an instant without ever turning around. That is why classification (the derivative tests) is a separate, necessary step after you find the candidates. The connection between a zero-slope point and its horizontal tangent line is the geometric heart of the whole method.
Who Discovered Critical Points?
The method for finding maxima and minima is older than calculus itself. Long before anyone wrote $\frac{d}{dx}$, a French lawyer solving problems for fun found a way to locate the high and low points of a curve.
Two later figures gave the idea its modern machinery:
Isaac Newton (1642–1727, England) and Gottfried Wilhelm Leibniz (1646–1716, Germany) independently built the derivative into a full system of calculus in the late 1600s, turning Fermat's trick into a general tool. Their bitter priority dispute over who invented calculus first split European mathematics for a century.
Joseph-Louis Lagrange (1736–1813, Italy and France) introduced the prime notation $f'(x)$ that this article uses, and formalised the theory of maxima and minima that Fermat had started.
Where Are Critical Points Used In The Real World?
The same idea, find where the rate of change is zero, quietly drives optimization across many fields.
Physics and motion: the highest point of a thrown ball is where its vertical velocity (the derivative of height) is zero, a critical point of the height function.
Economics and business: profit is maximised where marginal profit (the derivative of the profit function) is zero, which is exactly a critical point of profit.
Engineering design: minimising the material in a can or the drag on a wing means minimising a function, so engineers hunt for its critical points.
Machine learning: training a model means minimising a loss function, and the algorithm searches for points where the gradient (the multivariable derivative) is zero.
Biology and medicine: peak drug concentration in the bloodstream, or the fastest moment of population growth, are critical points of the concentration or population curve.
One habit of mind, locate the flat spots, turns up wherever something needs to be made as large or as small as possible.
What Are The Most Common Mistakes With Critical Points?
These four errors account for most lost marks on critical-point problems, and each one is a documented, recurring question from real students on r/learnmath, r/calculus, and math forums.
Forgetting the points where the derivative is undefined.
Where it slips in:
A student computes $f'(x)$, solves $f'(x) = 0$, writes down those answers, and stops, missing corners and cusps entirely.
Don't do this:
Do not treat "set the derivative to zero" as the whole search. That step never finds a point like $x = 0$ for $f(x) = x^{2/3}$.
The correct way:
Always run both checks. Solve $f'(x) = 0$, then separately find every $x$ where $f'(x)$ is undefined but $f(x)$ is defined. Both kinds are critical points.
Assuming every critical point is a maximum or minimum.
Where it slips in:
A student finds a critical point and immediately labels it a peak or a valley without testing it.
Don't do this:
Do not skip the classification step. For $f(x) = x^3$, the point $x = 0$ is critical but is neither a maximum nor a minimum.
The correct way:
Classify every critical point with the first or second derivative test. A critical point is only a candidate until a test confirms what it is.
Treating endpoints as critical points.
Where it slips in:
While hunting for the absolute maximum on a closed interval, a student lumps the endpoints in with the critical points and calls them all "critical."
Don't do this:
Do not label the endpoints of an interval as critical points. Critical points are interior points where $f'$ is zero or undefined.
The correct way:
Keep two separate lists for an absolute-extremum problem: the interior critical points, and the endpoints. Compare the function values across both lists to find the absolute maximum and minimum.
Misusing the second derivative test when $f''(c) = 0$.
Where it slips in:
A student computes $f''(c) = 0$ and concludes the point is "not an extremum," treating an inconclusive result as an answer.
Don't do this:
Do not read $f''(c) = 0$ as "neither a max nor a min." The test has simply failed to decide.
The correct way:
When $f''(c) = 0$, switch to the first derivative test and check the sign of $f'$ on each side of $c$. That test always gives a verdict.
Practice Problems On Critical Points
Find (and where asked, classify) the critical points. Answers follow each problem.
$f(x) = x^2 - 4x + 3$.
(Answer: $f'(x) = 2x - 4 = 0$ gives $x = 2$; $f''(2) = 2 > 0$, a local minimum at $(2, -1)$.)$f(x) = x^3 - 6x^2 + 9x$.
(Answer: $f'(x) = 3(x - 1)(x - 3)$, so $x = 1$ and $x = 3$; local maximum at $(1, 4)$, local minimum at $(3, 0)$.)$f(x) = 2x^3 - 3x^2 - 12x + 5$.
(Answer: $f'(x) = 6(x - 2)(x + 1)$, so $x = -1$ and $x = 2$; local maximum at $(-1, 12)$, local minimum at $(2, -15)$.)$f(x) = x^{1/3}$.
(Answer: $f'(x) = \frac{1}{3\sqrt[3]{x^2}}$ is never zero and is undefined at $x = 0$, which is in the domain, so $x = 0$ is the only critical point. Since $f'(x) > 0$ on both sides, it is not an extremum, it is a vertical-tangent inflection.)$f(x) = x,e^{-x}$.
(Answer: $f'(x) = e^{-x}(1 - x) = 0$ gives $x = 1$; $f''(1) = -e^{-1} < 0$, a local maximum at $\left(1, \tfrac{1}{e}\right) \approx (1, 0.3679)$.)
Where Should You Go Next After Critical Points?
Critical points sit at the centre of the applications of derivatives, and several natural doors open from here.
Increasing And Decreasing Intervals. The first derivative test is really a question about where a function rises and falls, and this is where that sign analysis is built from the ground up.
Tangents And Normals. A critical point is defined by its tangent line, so understanding tangents (and the normals perpendicular to them) deepens the geometry behind the rule.
Integration. After differentiation and its applications, integration is the other half of calculus, the accumulation and area side of the story.
If your child is learning to find and classify critical points, a live Bhanzu trainer teaches the topic starting from the "why" (Fermat's flat-tangent idea and the geometry of peaks and valleys) in the Bhanzu math classes.
Was this article helpful?
Your feedback helps us write better content