Derivatives · Lesson 3 of 4 · ~6 min

Critical points

Predict

The slope function of $f(x) = x^3/3 - x$ is $f'(x) = x^2 - 1$, which crosses zero at $x = -1$ going from positive to negative. What does $f$ have at $x = -1$?

Hold that thought. Drag x slowly across $-1$ in the instrument and watch the state readout.

Try

Try itPress x = −1: the tangent segment goes flat and f′(x) reads $0.000$. Then drag x slowly across $-1$ and across $+1$, watching the state readout. The dotted band along the bottom is the sign chart of $f'$.

Notice

Approaching $-1$ from the left the state reads rise; at $-1$ it flips to local max; past it, fall. The same sweep across $x = 1$ runs fall, local min, rise. Both turning points sit exactly where f′(x) reads $0.000$ — and nowhere else on the curve does the direction change.

Explain

Peaks and valleys cannot happen where the slope is nonzero: a positive slope offers a higher point one step right, a negative slope one step left. So the only candidates are the points where that argument breaks down, and they get a name — a critical point of $f$ is a $c$ in the domain where

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

Here $f'(x) = x^2 - 1 = (x-1)(x+1)$ gives exactly two candidates, $x = -1$ and $x = 1$. A candidate is not a verdict. The First Derivative Test delivers one by reading the sign chart: if $f'$ switches from positive to negative at $c$, then $f$ rises into $c$ and falls out of it — a local maximum. Switching negative to positive gives a local minimum, and no switch gives neither.

Apply

Checkpoint — prove it to yourself

Sweeping across $x = 1$, the sign chart flips from $f' < 0$ to $f' > 0$. What does the First Derivative Test conclude there?

Press x = 0. The readout shows $f'(0) = -1.000$. Is $x = 0$ a critical point of this $f$?

Transfer: $g(x) = x^3$ has $g'(x) = 3x^2$, so $g'(0) = 0$ — but $g' > 0$ on both sides of $0$. What is $x = 0$ for $g$?

Key idea

Critical points are where extremes can happen; the sign switch of $f'$ decides whether they do. Zero slope with no sign change is just a pause mid-climb — candidate, then test, always in that order.

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