Increasing and Decreasing Functions

Falling, then rising

A function is increasing on an interval if, wherever $x_1<x_2$x₁ < x₂ in that interval, $f\left(x_1\right)<f\left(x_2\right)$f(x₁) < f(x₂): moving right, the graph goes up. It is decreasing if moving right the graph goes down.

On $y=x^2$y = x² 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 $f\left(x\right)=x^3-3x$f(x) = x³ − 3x, $f^{\prime }\left(x\right)=3x^2-3=3\left(x-1\right)\left(x+1\right)$f'(x) = 3x² − 3 = 3(x − 1)(x + 1), 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 $x^3-3x$x³ − 3x 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 $y=x^3-3x$y = x³ − 3x and, softer, its derivative $y=3x^2-3$y = 3x² − 3. 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 $f\left(x\right)=2x^3-9x^2+12x$f(x) = 2x³ − 9x² + 12x, $f^{\prime }\left(x\right)=6x^2-18x+12=6\left(x-1\right)\left(x-2\right)$f'(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2), 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 $f\left(x\right)=x^3$f(x) = x³, $f^{\prime }\left(x\right)=3x^2$f'(x) = 3x² is 0 at x = 0 and positive on both sides. The curve is level for an instant at the origin and goes on climbing: $x^3$x³ 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 $f\left(x\right)=\frac{1}{x}$f(x) = 1/x, $f^{\prime }\left(x\right)=-\frac{1}{x^2}$f'(x) = −1/x², which is negative wherever it exists. So $\frac{1}{x}$1/x 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.

The curve $y=\frac{x^3}{3}-x$y = x³/3 − x, whose derivative is $x^2-1$x² − 1. 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: $x^3-3x$x³ − 3x 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 $x^3-3x$x³ − 3x, f' = 0 says nothing about the intervals on either side; test a value inside each interval instead.

Joining intervals across a gap. $\frac{1}{x}$1/x decreases on x < 0 and on x > 0, but not on the two together.