Stationary Points: Maxima, Minima & Inflection

What Are Stationary Points?

Stationary points are the points on the graph of a function where the derivative equals zero:

$$f'(x) = 0$$

Because the derivative measures the slope of the tangent line, $f'(x) = 0$ means the tangent is horizontal at that point. The curve is momentarily neither rising nor falling. Geometrically, that is the "flat spot" on a graph: the peak of a hill, the bottom of a valley, or a brief level pause on an otherwise climbing curve.

Every stationary point falls into one of exactly three types:

• Local maximum: the curve rises up to the point and falls away after it, a peak.

• Local minimum: the curve falls down to the point and rises away after it, a valley bottom.

• Point of inflection (stationary or horizontal inflection): the curve flattens for an instant but keeps heading the same overall direction, so it is neither a peak nor a valley.

To find every stationary point of a function, differentiate it, set the derivative equal to zero, and solve for $x$. Each solution is the $x$-coordinate of a stationary point; substituting it back into $f(x)$ gives the height. What is left is to decide which of the three types each one is, which is what the two classification tests below are for. The derivative itself comes from the rules of differentiation, so those need to be solid first.

Are Stationary Points The Same As Critical Points?

Not quite, and this is where many students trip. A critical point is a point in the domain where the derivative is either zero or undefined. Stationary points are the first kind only.

$$\text{Stationary point: } f'(x) = 0 \qquad \text{Critical point: } f'(x) = 0 \ \text{ or } \ f'(x) \text{ does not exist}$$

So every stationary point is a critical point, but not every critical point is stationary. The corner of $f(x) = |x|$ at $x = 0$ is a critical point (the derivative does not exist there) yet it is not a stationary point, because there is no horizontal tangent. Stationary points are the smooth, horizontal-tangent subset of the critical points. Keeping the two words apart saves a surprising number of lost marks.

How Do You Classify A Stationary Point With The Second-Derivative Test?

Once you have a stationary point at $x = a$ (so $f'(a) = 0$), the fastest way to name it is usually the second-derivative test. It reads the concavity of the curve at that point.

$$\begin{aligned} f''(a) &< 0 \quad\Rightarrow\quad \text{local maximum (curve bends downward, a cap)}\ f''(a) &> 0 \quad\Rightarrow\quad \text{local minimum (curve bends upward, a cup)}\ f''(a) &= 0 \quad\Rightarrow\quad \text{inconclusive, the test fails, check further} \end{aligned}$$

The geometry behind it is direct. If the curve is concave down at the flat spot, it must be sitting at a peak; if it is concave up, it must be sitting in a valley. When $f''(a) = 0$ the curve is not committing to either bend at that instant, so the test gives no verdict and you fall back on the first-derivative test. For the full treatment of the concavity idea, see the second derivative and the dedicated second derivative test.

How Do You Classify A Stationary Point With The First-Derivative Test?

The first-derivative test looks at the sign of $f'$ just before and just after the stationary point. It always works, including in the case where the second-derivative test fails.

Table: Reading the sign of $f'$ on each side of a stationary point at $x = a$.

The rule in words: if $f'$ changes sign at the point, it is a turning point (a maximum or a minimum); if $f'$ keeps the same sign on both sides, it is a horizontal point of inflection. This is the whole reason a horizontal inflection is not an extremum, the curve pauses but never actually turns around. The test connects directly to where a function is increasing and decreasing, and it is set out in full at the first derivative test.

How Do You Find And Classify Stationary Points? Worked Examples

Each example finds every stationary point by solving $f'(x) = 0$, then names it. The examples are chosen to show all three outcomes, including the two cases where $f''(x) = 0$.

Example 1: A cubic with a maximum and a minimum.

Classify the stationary points of $f(x) = x^3 - 3x$.

Differentiate and set the derivative to zero:

$$f'(x) = 3x^2 - 3 = 0 \quad\Rightarrow\quad x^2 = 1 \quad\Rightarrow\quad x = \pm 1$$

So there are stationary points at $x = -1$ and $x = 1$. Use the second derivative $f''(x) = 6x$:

$$f''(-1) = -6 < 0 \ \Rightarrow\ \text{maximum}, \qquad f''(1) = 6 > 0 \ \Rightarrow\ \text{minimum}$$

The heights are $f(-1) = (-1)^3 - 3(-1) = 2$ and $f(1) = (1)^3 - 3(1) = -2$.

Final answer: local maximum at $(-1, 2)$, local minimum at $(1, -2)$.

Example 2: A stationary point that is not a maximum or minimum.

Classify the stationary point of $f(x) = x^3$.

$$f'(x) = 3x^2 = 0 \quad\Rightarrow\quad x = 0$$

There is one stationary point, at $x = 0$. The second-derivative test gives $f''(x) = 6x$, so $f''(0) = 0$: inconclusive. Switch to the first-derivative test. Since $f'(x) = 3x^2$ is positive on both sides of $0$ (it is a square), the slope never changes sign, so the curve keeps rising through the flat spot.

Final answer: $x = 0$ is a horizontal point of inflection, not a maximum or a minimum.

Example 3: When $f''(x) = 0$ but the point still is an extremum.

Classify the stationary point of $f(x) = x^4$.

$$f'(x) = 4x^3 = 0 \quad\Rightarrow\quad x = 0$$

Again $f''(x) = 12x^2$ gives $f''(0) = 0$, so the second-derivative test fails a second time. Do not conclude "inflection" from that alone. Apply the first-derivative test: $f'(x) = 4x^3$ is negative for $x < 0$ and positive for $x > 0$, so the slope changes from falling to rising.

Final answer: $x = 0$ is a local minimum, at the point $(0, 0)$, even though $f''(0) = 0$.

Examples 2 and 3 together carry the single most important lesson of this topic: $f''(x) = 0$ decides nothing on its own. One such point was an inflection, the other a genuine minimum, and only the first-derivative test told them apart.

What Are The Three Types Of Stationary Points? A Reference Table

Table: The three stationary points, their derivative signatures, and a standard example.

The table also makes the trap visible: $f''(a) = 0$ appears on only one row, but as Example 3 showed, an $f''(a) = 0$ result can still hide a maximum or a minimum. The value $f''(a) = 0$ is a signal to check the sign pattern, never a conclusion by itself.

Why Do Stationary Points Work The Way They Do?

The whole subject rests on one geometric fact: the derivative is the slope of the tangent line, so $f'(x) = 0$ is exactly the statement that the tangent is flat.

• A flat tangent is where direction can change. A smooth curve can only switch from rising to falling (or the reverse) by passing through a moment of zero slope. That is why local maxima and minima, the turning points, must be stationary points.

• Flat does not force a turn. A curve can flatten for an instant and then carry on in the same direction, as $x^3$ does at the origin. The slope touches zero without changing sign, giving a horizontal inflection rather than a turn.

• Concavity is what the second derivative reads. $f''(x)$ measures how the slope itself is changing. A negative $f''$ means the slope is decreasing (the curve caps over into a maximum); a positive $f''$ means the slope is increasing (the curve cups up into a minimum). When $f''$ is zero, the concavity is not committing, so the test stays silent.

Read together, the two tests are the same idea from two angles. The first-derivative test watches the slope cross zero; the second-derivative test watches how fast the slope is turning as it does. Both are downstream of the applications of derivatives that treat the derivative as a rate of change.

Who Shaped The Study Of Stationary Points?

The idea of using a "flat tangent" to locate a maximum or a minimum is older than calculus itself.

Two points on the wider history are worth naming:

• Pierre de Fermat (1601–1665, France) gave the condition for a maximum or minimum, the seed of the modern derivative test.

• Joseph-Louis Lagrange (1736–1813, Italy and France) later helped formalise the distinction between stationary points and true extrema, sharpening the logic that a vanishing derivative is necessary but not sufficient for a maximum or minimum.

Where Are Stationary Points Used In The Real World?

Anywhere a quantity is pushed to its highest or lowest value, a stationary point is doing the work behind the scenes.

• Projectile motion: the top of a ball's flight, where vertical speed is momentarily zero, is a stationary point of its height-versus-time curve. Setting $f'(t) = 0$ finds the exact time and height of the apex.

• Business and economics: a company maximises profit or minimises cost by differentiating the profit or cost function and solving $f'(x) = 0$, then confirming a maximum or minimum with the second-derivative test.

• Engineering and design: finding the least material for a container of fixed volume, or the strongest beam for a given weight, reduces to locating a minimum, a stationary point of the relevant function.

• Physics and equilibrium: a system settles where its potential energy is stationary; a stable equilibrium sits at a local minimum, an unstable one at a local maximum.

• Data and optimisation: algorithms that fit models search for stationary points of an error function, driving the slope toward zero to land at the best fit.

These are all the same move in different clothing: to optimise a quantity, find where its rate of change is zero, then classify what kind of stationary point you have found. The topic is the front door to the optimization problems that fill applied calculus.

What Are The Most Common Mistakes With Stationary Points?

These three errors account for most lost marks on stationary points, and each matches a question real students ask on r/learnmath, r/calculus, and university common-error handouts.

Assuming every stationary point is a maximum or minimum.

Where it slips in:

A student solves $f'(x) = 0$, finds a point, and immediately calls it a turning point without checking the type.

Don't do this:

Do not treat $f'(x) = 0$ as proof of a peak or a valley. A stationary point can also be a horizontal point of inflection, as $f(x) = x^3$ is at $x = 0$.

The correct way:

Always classify. Use the first-derivative test (does $f'$ change sign?) or the second-derivative test. Only a genuine sign change of $f'$ makes a point a maximum or minimum.

Stopping at $f''(x) = 0$ and calling it an inflection.

Where it slips in:

The second-derivative test returns $f''(a) = 0$, and a student concludes the point must be a point of inflection.

Don't do this:

Do not read $f''(a) = 0$ as a verdict. It only means the test is inconclusive. The point could be a maximum, a minimum, or an inflection.

The correct way:

Fall back on the first-derivative test. For $f(x) = x^4$, $f''(0) = 0$, yet the sign of $f'$ changes from negative to positive, so it is a minimum, not an inflection.

Confusing stationary points with all critical points.

Where it slips in:

A student is asked for stationary points and includes a corner or a cusp where the derivative does not exist, such as $x = 0$ on $f(x) = |x|$.

Don't do this:

Do not fold "derivative undefined" points into stationary points. Those are critical points but have no horizontal tangent.

The correct way:

A stationary point needs $f'(x) = 0$ exactly. Points where $f'$ fails to exist are critical but not stationary, keep the two categories separate.

Practice Problems On Stationary Points

Work each one, then check against the answer. Every answer is verified.

1. Find the stationary point of $f(x) = x^2 - 4x + 3$ and classify it.
   (Answer: $f'(x) = 2x - 4 = 0$ gives $x = 2$; $f''(x) = 2 > 0$, so a minimum at $(2, -1)$.)

2. Classify the stationary points of $f(x) = x^3 - 3x$.
   (Answer: maximum at $(-1, 2)$, minimum at $(1, -2)$.)

3. Find and classify the stationary points of $f(x) = x^3 - 6x^2 + 9x$.
   (Answer: $f'(x) = 3x^2 - 12x + 9 = 0$ gives $x = 1, 3$; maximum at $(1, 4)$, minimum at $(3, 0)$.)

4. Find and classify the stationary points of $f(x) = x^4 - 2x^2$.
   (Answer: $x = 0$ maximum at $(0, 0)$; $x = -1$ and $x = 1$ each a minimum with value $-1$.)

5. Show that $f(x) = x^5$ has a stationary point at $x = 0$ that is not a maximum or minimum.
   (Answer: $f'(x) = 5x^4 = 0$ at $x = 0$; $f'$ stays positive on both sides, so it is a horizontal point of inflection.)

6. Find and classify the stationary points of $f(x) = 2x^3 - 3x^2 - 12x + 5$.
   (Answer: $f'(x) = 6x^2 - 6x - 12 = 0$ gives $x = -1, 2$; maximum at $(-1, 12)$, minimum at $(2, -15)$.)

Where Should You Go Next After Stationary Points?

Stationary points are the hinge of applied differentiation, and several natural doors open from here.

1. Second derivative test. Master the concavity test you used to classify each point, including exactly when it fails.

2. Curve sketching. Combine stationary points with intercepts and asymptotes to draw a full graph by hand.

3. Optimization problems. Turn the classification skill into real maximum-and-minimum word problems in geometry, physics, and economics.

If your child is meeting stationary points for the first time, a live Bhanzu trainer teaches them from the horizontal-tangent picture up, so finding and classifying feel like one connected idea, through the Bhanzu math program.