Critical Points

Throw a ball straight up. For a moment it climbs, then — at the very top of its flight — it hangs, perfectly still, before falling back. A company raises its price and revenue climbs… until it doesn't: push the price past a certain point and customers walk away, so somewhere in between sits the price that earns the most. A factory finds that making more units spreads its fixed costs thinner, but only up to a point — past it, overtime and wear drive the average cost back up, and somewhere in the middle is the cheapest output level.

Three completely different stories, one mathematical event: a quantity that was rising stops rising (or a falling one stops falling) and turns around. The top of the ball's arc, the profit-maximising price, the cost-minimising output — every one of them lives at a special input of a function. This page is about finding those special inputs. They are called critical points, and they are the master key to every optimisation problem you will ever meet.

The definition

The interesting places on a curve — the peaks, the valleys, the moments it pauses — all happen where the slope stops being uphill or downhill. A critical point is an input x = c in the domain of f where

f'(c) = 0 \qquad\text{or}\qquad f'(c)\ \text{does not exist.}

Both halves matter. A smooth curve can only turn around where its tangent goes flat — that is the f'(c) = 0 case. But a curve with a sharp corner or a vertical cusp can turn around without ever having a flat tangent, because at the turning instant it has no well-defined slope at all — that is the "does not exist" case. And the quiet phrase "in the domain" is doing real work too: if f isn't even defined at c, there is nothing there to be critical.

Why do we care? Because critical points are the only candidates for turning. We already know that the sign of f' tells us where f rises and falls — and the sign of a derivative can only switch at a place where the derivative is zero or undefined. Critical points are the fence posts between the rising stretches and the falling stretches; between two consecutive ones, the function is committed to a single direction.

If f has a local maximum or local minimum at an interior point c of its domain, and f is differentiable at c, then f'(c) = 0.
Consequently, every local maximum or minimum of f at an interior point occurs at a critical point — a point where f' is zero or fails to exist.

Read the theorem's direction carefully: it says peaks and valleys must be critical points, not that critical points must be peaks or valleys. That one-way arrow is the source of the classic mistakes on this topic — we'll ambush them below.

Watch the tangent go flat

Slide the point along f(x) = x^3 - 3x and watch its tangent line. Almost everywhere the tangent tilts: steeply up on the far left, down through the middle, steeply up again on the right. But at exactly two inputs — x = -1 and x = 1 — it goes perfectly horizontal: slope zero. Those two inputs are the critical points.

Notice what the curve does around each one. Just left of x = -1 the curve rises; just right of it the curve falls — so x = -1 is the top of a hill, a local maximum. Around x = 1 the pattern flips — falling, then rising — the bottom of a valley, a local minimum. The slider's snap points land on both, so you can park the tangent exactly on each flat spot.

Worked example 1: a cubic, start to finish

Find all critical points of f(x) = 2x^3 - 9x^2 + 12x.

Step 1 — differentiate.

f'(x) = 6x^2 - 18x + 12.

Step 2 — where is f' undefined? Nowhere: f' is a polynomial, defined for every real x. So for this function only the f'(x) = 0 case can fire. (Always ask this question anyway — it takes two seconds and it is the half of the definition everyone forgets.)

Step 3 — solve f'(x) = 0. Take out the common factor first; derivatives of polynomials almost always factor nicely in a set exercise.

6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2) = 0 \;\Rightarrow\; x = 1 \ \text{or}\ x = 2.

Step 4 — check they're in the domain. The domain of a polynomial is all of \mathbb{R}, so both survive. The critical points are x = 1 and x = 2, and the points on the curve are (1, f(1)) = (1, 5) and (2, f(2)) = (2, 4).

The same recipe cracked the curve in the chart above: f(x) = x^3 - 3x gives f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0, so x = \pm 1 — exactly where you saw the tangent flatten.

Worked example 2: when f' doesn't exist

Find all critical points of f(x) = x^{2/3}.

Step 1 — differentiate. By the power rule,

f'(x) = \tfrac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}.

Step 2 — solve f'(x) = 0. A fraction is zero only when its numerator is zero, and the numerator here is the constant 2 — never zero. So there are no solutions of f'(x) = 0 at all.

Step 3 — where is f' undefined? At x = 0, where the denominator vanishes. And crucially, x = 0 is in the domain of f itself: f(0) = 0 is perfectly fine. Undefined derivative, defined function — that is exactly the second half of the definition. So x = 0 is a critical point, and it is the only one.

Look at the graph and you'll see why the derivative gives up there: the curve comes down into the origin ever more steeply, turns on a knife-edge, and climbs out just as steeply. The tangent lines approach vertical from both sides — a shape called a cusp. And the cusp is not a technicality: it is precisely where the minimum of the function lives. Skip the "undefined" check and you'd miss the minimum entirely, because f'(x) = 0 has no solutions here.

The absolute-value function pulls the same trick more gently: f(x) = |x| has slope -1 everywhere left of zero and +1 everywhere right of it, so the slope never equals zero — yet the graph has an unmistakable minimum at the corner x = 0, where f' doesn't exist.

Two kinds of critical point, side by side

So a critical point can occur for two reasons: the slope is zero (a smooth peak, valley, or momentary flat) or the slope is undefined (a sharp corner or a cusp, where the curve changes direction without ever levelling out). Step through both below — the smooth valley has a horizontal dashed tangent; the corner has no tangent to draw at all.

In an exam, the zero-slope kind dominates — most functions you'll be handed are polynomials or other smooth curves. But the moment a modulus sign, a fractional power, or a piecewise definition appears, put the "where is f' undefined?" check at the top of your list.

Trap 1: a critical point is not automatically a max or a min.

For g(x) = x^3 we get g'(x) = 3x^2 = 0 at x = 0 — a genuine critical point. But the curve only flattens for an instant and keeps climbing: g is increasing on both sides, so x = 0 is neither a maximum nor a minimum. Fermat's theorem is a one-way street: extrema \Rightarrow critical points, but critical points \not\Rightarrow extrema. Finding critical points only gives you the candidates; classifying them (next section) is a separate job.

Trap 2: forgetting the "does not exist" half.

"Solve f'(x) = 0" is only half the definition. As |x| and x^{2/3} show, a function can put its minimum at a point where the derivative fails to exist — and there f'(x) = 0 has no solutions to find. One refinement, though: the point must be in the domain of f. For f(x) = 1/x, the derivative -1/x^2 is undefined at x = 0 — but so is f itself, so x = 0 is not a critical point. No point of the graph, no critical point.

Worked example 3: classify what you found

Critical points in hand, deciding what each one is takes only the sign of f' on either side. The logic is the picture from the tangent chart, made systematic: if f switches from rising to falling across c, then c is a local maximum; falling to rising, a local minimum; no switch — as with x^3 — neither.

Take f(x) = x^3 - 3x again, with f'(x) = 3(x-1)(x+1) and critical points x = \pm 1. The two critical points slice the number line into three intervals, and on each interval f' keeps one sign — so one test value per interval settles everything. Step through the chart:

Conclusion: a local maximum at (-1, 2) and a local minimum at (1, -2) — precisely the hilltop and valley you slid the tangent over earlier. This "check the sign of f' either side" procedure is the first derivative test, and it works even at corners and cusps, where fancier tools (like the second derivative) have nothing to say. Try it on x^{2/3}: the derivative \tfrac{2}{3}x^{-1/3} is negative for x < 0 and positive for x > 0 — fall then rise, so the cusp is a minimum, exactly as the graph shows.

Pierre de Fermat — a full generation before Newton or Leibniz wrote down a derivative — had a method for finding maxima that is eerily close to the modern one. He called it adequality. To maximise an expression f(x), Fermat compared f(x) with f(x + e) for a tiny nudge e, and reasoned: near a peak, a small step barely changes the value, so treat the two as "adequal", f(x+e) \approx f(x). Expand, cancel, divide through by e — and then, in the step that scandalised his critics, simply set e = 0. Having just divided by it! It worked flawlessly: dividing a segment of length b to maximise x(b - x), his method gives b - 2x = 0, i.e. cut it in half — which is exactly f'(x) = 0 in disguise, computed decades before derivatives had a name or a notation.