Mathematics · Study Guide

When Is a Function Decreasing? Solve It with Derivatives

A curve is decreasing on an interval when its y-values fall as x moves from left to right. In the usual A-Level method, find the derivative, then solve where it is negative. The crucial distinction is between the curve lying below the x-axis and the curve moving downhill: those describe different things.

Turn the words into a mathematical test

To find open intervals where the gradient is negative, solve $dy/dx<0$ . The inequality concerns the derivative, not y itself. A curve can be above the x-axis and still be decreasing. It can also be below the axis while increasing. Draw a short downhill line above an axis if you need to separate the two ideas. The position of the graph and the sign of its slope are independent.

Find the boundaries before testing intervals

Differentiate first, then solve $dy/dx=0$ to find stationary x-values. Also note any points where the function or derivative is undefined; those can split the domain into further intervals. Between consecutive roots of a continuous derivative, its sign cannot change without passing through zero. That is why one test value in each such interval is enough for the polynomial example below.

Read a sign chart correctly

Put the boundary values in increasing order on a number line. Choose one x-value from each interval and substitute into the derivative. Record positive, negative or zero. For a product of two factors, check the sign of each factor, then multiply the signs. Do not pick a test value equal to a boundary: that gives zero and tells you nothing about either neighbouring interval.

Why we use strict inequalities in the answer

At a stationary boundary, the gradient is zero, so that x-value is not a solution of $dy/dx<0$ . For the standard exam question asking where the gradient is negative, give open intervals. The broader mathematical definition of a strictly decreasing function can still include an isolated point with zero derivative, as with $y=-x^3$ . Do not conclude that every zero derivative must mark a change from increasing to decreasing.

Connect the algebra back to the original curve

After solving the inequality, describe the movement in words. In the example, the original cubic rises before x = −1/2, falls between −1/2 and 4, then rises again. The diagram shows the derivative, not the cubic. Its section below zero identifies where the original curve has a negative tangent gradient. Keeping those two graphs separate helps prevent a common interpretation error.

Worked example: a cubic curve

Question

Find where

y=(2/3)x^3-(7/2)x^2-4x+5

is decreasing.

Differentiate

dy/dx=2x^2-7x-4

. The constant term contributes zero.

Factorise

2x^2-7x-4=(2x+1)(x-4)

. Its zeros are

x=-1/2

and

x=4

Test signs

At x = −1 the derivative is 5, which is positive. At x = 0 it is −4, which is negative. At x = 5 it is 11, which is positive.

Conclude

The derivative is negative only between the roots. Therefore the required interval is

-1/2<x<4

The derivative 2x squared minus 7x minus 4 crosses zero at minus one half and four. It is negative between those roots and positive outside. — This is the derivative graph, $y\prime=2x^2-7x-4$ . The highlighted region shows where the original cubic decreases.

Common Mistakes

Solving y < 0 instead of dy/dx < 0.
Reporting only the stationary x-values instead of the interval between them.
Assuming a zero derivative always means a maximum or minimum.
Including endpoints even though the question asks for a strictly negative gradient.

Practice

Where does

y=x^2-6x+2

have negative gradient?

dy/dx=2x-6<0

, so x < 3.

Where does

y=x^3-3x

have negative gradient?

dy/dx=3x^2-3=3(x-1)(x+1)

, which is negative for −1 < x < 1.

y=x^3

decreasing for negative x?

No. Its derivative is

3x^2

, which is positive when x is negative. Negative y-values do not imply a decreasing curve.

Put it into practice

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