The sign of the derivative says whether a function is increasing or decreasing
Falling, then rising
A function is increasing on an interval if, wherever in that interval, : moving right, the graph goes up. It is decreasing if moving right the graph goes down.
On the derivative is 2x. Left of the minimum, for x < 0, the derivative is negative, at x = −2 it is −4, and the function is decreasing: from x = −2 to x = −1 it drops from 4 to 1. Right of the minimum, for x > 0, the derivative is positive and the function is increasing: from x = 1 to x = 2 it climbs from 1 to 4.
The sign decides
The derivative is the gradient of the tangent. A positive gradient means the curve is climbing at that point, and a negative one means it is falling. So f is increasing on an interval where f′(x) > 0 throughout, and decreasing where f′(x) < 0 throughout.
Only the sign matters, not the size. A derivative of 0.2 is small, but it is positive, so the function is increasing there, gently. A derivative of −3 means falling, whatever the sign of x is.
Finding the intervals
To find where f increases and decreases: differentiate, solve f′(x) = 0, mark the solutions on a number line, and test the sign of f′ at one value in each interval between them.
For , , which is 0 at x = −1 and x = 1. Those split the line into three intervals. At x = −2, f' = 12 − 3 = 9, positive. At x = 0, f' = −3, negative. At x = 2, f' = 9, positive.
So is increasing for x < −1, decreasing for −1 < x < 1, and increasing for x > 1. The heights agree: f(−1) = −1 + 3 = 2 and f(1) = 1 − 3 = −2, so between them the function falls from 2 to −2.
One test value is enough for each interval. f′ is a polynomial, so it changes continuously, and it could only change sign by passing through 0; between two neighboring zeros it keeps one sign.
The curve and, softer, its derivative . Where the derivative is above the x-axis, for x < −1 and x > 1, the curve rises; between −1 and 1 the derivative is below the axis and the curve falls from (−1, 2) to (1, −2).
A second example
For , , which is 0 at x = 1 and x = 2. Test one value in each interval: f'(0) = 12, positive; f'(1.5) = 6 × 0.5 × (−0.5) = −1.5, negative; f'(3) = 12, positive.
So f is increasing for x < 1, decreasing for 1 < x < 2, and increasing for x > 2. The heights confirm it: f(1) = 2 − 9 + 12 = 5 and f(2) = 16 − 36 + 24 = 4, a drop of 1 between them.
Zero without a change of sign
A zero of f' does not always split the line into a rise and a fall. For , is 0 at x = 0 and positive on both sides. The curve is level for an instant at the origin and goes on climbing: is increasing everywhere.
So it is the sign of f′ on each interval that gives the answer, not the zeros themselves.
On intervals, not across gaps
For , , which is negative wherever it exists. So is decreasing for x < 0 and decreasing for x > 0. It is not decreasing across 0: f(−1) = −1 is less than f(1) = 1. The function is not defined at 0, and the statement holds on each of the two intervals separately.
Drag P to a turning point and watch the normal
The curve , whose derivative is . At x = −1.5 the tangent has gradient m = 1.25, positive, and the curve is rising. Drag the point P between −1 and 1: m turns negative and the curve is falling; past 1 it is positive again.
The usual mistakes
Reading the sign of x instead of the sign of f'. A curve can climb where x is negative: is increasing at x = −2, where f' = 9.
Treating a small derivative as no direction. f′ = 0.2 is positive, so the function is increasing there.
Testing at the zeros of f'. At x = 1 on , f' = 0 says nothing about the intervals on either side; test a value inside each interval instead.
Joining intervals across a gap. decreases on x < 0 and on x > 0, but not on the two together.
Your turn
1.f′(x) = 0.2 at a point. Is f increasing there?