Every graph tells a story, and you read its mood at a glance. A share price climbing all morning — good mood. A mug of coffee cooling on your desk — temperature sliding down. A bathtub filling — water level rising steadily until someone turns off the tap. In each case the question that matters is not "how big is the value right now?" but "which way is it heading?"

The derivative answers exactly that question, because f'(x) is the slope of the curve at x — and a slope points either uphill or downhill. The sign of f' is the story:

• If f'(x) > 0, the slope points uphill — f is increasing: its graph rises as you move to the right.
• If f'(x) < 0, the slope points downhill — f is decreasing: its graph falls as you move to the right.
• If f'(x) = 0, the slope is flat for an instant — the curve is pausing, and possibly (though not always!) turning around.

Notice that you never need to see the graph of f to know its mood. Compute f', check its sign, and you know whether the original curve is rising or falling — even for a function far too messy to sketch. That's the trick this page teaches: reading the shape of a function straight off the sign of its derivative. A function that only ever does one of the two — always rising, or always falling — is called monotonic, and we'll meet a famous one below.

Watch the sign of the slope

Here is f(x) = x^3 - 3x in bold, with its derivative f'(x) = 3x^2 - 3 drawn dashed on the same axes. Slide the marker and watch the tangent line: when the tangent tilts uphill, the dashed f' curve sits above the x-axis (positive); when the tangent tilts downhill, f' sits below (negative).

Pay special attention to the two moments where the bold curve turns around — the hilltop near x = -1 and the valley floor near x = 1. At exactly those x-values the dashed curve crosses zero. That is no coincidence: a smooth function can only switch from rising to falling (or back) where its slope passes through 0. So the roots of f' are the candidate turning points of f — which is why every analysis on this page starts by solving f'(x) = 0.

The sign line

Let's turn that observation into a method. To find where a function increases and decreases, first solve f'(x) = 0. For f'(x) = 3x^2 - 3 = 3(x-1)(x+1) that gives x = -1 and x = 1. These two roots chop the number line into three pieces, and here is the key fact: inside each piece the sign of f' cannot change (it would have to pass through zero to do so, and all the zeros are at the boundaries). So one test point per piece tells you the sign of f' on the whole piece:

So f increases, then decreases, then increases — exactly the N-shape you saw above, recovered without looking at the graph at all. The figure below makes the sign line interactive: drag the test point along the line and see whether f' there comes out positive or negative.

Worked example 1: a full sign-table analysis

Find the intervals on which f(x) = x^3 - 6x^2 + 9x + 1 is increasing and decreasing.

Step 1 — differentiate.

Step 2 — solve f'(x) = 0. Factor out the 3 first:

Step 3 — test one point in each interval. The roots split the line into three intervals. Pick easy test values — say x = 0, x = 2 and x = 4 — and track the sign of each factor in a sign table:

(Check the middle column with x = 2: f'(2) = 3(2-1)(2-3) = 3 \times 1 \times (-1) = -3 < 0. ✓)

Step 4 — conclude. f is increasing on x < 1 and on x > 3, and decreasing on 1 < x < 3. Without plotting a single point of f itself, we know its whole shape: up to a peak at x = 1, down to a trough at x = 3, then up forever. This four-step routine — differentiate, find the roots of f', test each interval, conclude — works on any polynomial you'll meet, and it's the backbone of curve sketching and of finding maxima and minima.

Worked example 2: a function that never falls

Where is f(x) = x^3 increasing?

Step 1: f'(x) = 3x^2. Step 2: f'(x) = 0 only at x = 0. Step 3: test both sides — f'(-1) = 3 > 0 and f'(1) = 3 > 0. A square is never negative, so f'(x) = 3x^2 \ge 0 everywhere, and it only touches zero at the single point x = 0.

Conclusion: x^3 is increasing on the entire real line. Look at the graph — the curve does go flat for one instant at the origin, but it never actually descends. Pick any two inputs with a < b and you will always find a^3 < b^3: the output never comes back down. One flat instant does not break an increasing run, just as a cyclist who stops pedalling for a single moment at the brow of a false flat is still, over the whole ride, always gaining altitude.

The dashed parabola is the derivative: it kisses the axis at x = 0 but never dips below it. Compare that with x^3 - 3x from earlier, whose derivative crossed through the axis and spent real time below it. Touching zero and crossing zero are different stories: only a genuine sign change of f' makes f turn around.

Worked example 3: reading a real story from T'

A mug of coffee is poured at 90\,^{\circ}\mathrm{C}. Over its first six minutes on the desk, its temperature is well modelled by

Is the coffee cooling the whole time? And when is it cooling fastest?

Step 1 — differentiate.

Step 2 — read the sign. On the model's domain t < 6, the factor (t - 6) is negative, so T'(t) < 0 throughout. Temperature falling ⇔ T' < 0 — the coffee is cooling for the entire six minutes, and we proved it without evaluating T once.

Step 3 — interpret the size, not just the sign. T'(0) = -12: at the start, the temperature drops at 12\,^{\circ}\mathrm{C} per minute. By t = 5, T'(5) = -2: still falling, but only at 2\,^{\circ}\mathrm{C} per minute. The sign of T' says which way; its magnitude says how fast. Hot coffee cools quickest when it's hottest — exactly what the shrinking steepness shows.

One caution about models: at t = 6 we get T'(6) = 0, and beyond that the parabola would have T' > 0 — the model would predict the coffee spontaneously reheating, which is nonsense. Checking the sign of the derivative is also how you spot where a model's validity runs out.

Watch it on Khan Academy