7.6 Increasing and decreasing functions — Learn

Prerequisites: differentiating polynomials from Section 7.3; solving linear and quadratic inequalities from Chapter 2; reading the sign of a factorised expression.

By the end of this section you can: decide where a function is increasing and where it is decreasing from the sign of its derivative; solve the inequalities $f'(x) > 0$ and $f'(x) < 0$ ; and link the intervals to the graph of the function.

1. The sign of the derivative

The derivative tells you which way a function is going.

On an interval where $f'(x) > 0$ , the function is increasing as $x$ increases.
On an interval where $f'(x) < 0$ , the function is decreasing as $x$ increases.

The word describes the direction of the curve, not its height. A function can be increasing while its values are negative, and decreasing while its values are positive.

To find the intervals, differentiate, then solve the inequalities $f'(x) > 0$ and $f'(x) < 0$ .

Example 1. Find where $f(x) = x^2 - 4x$ is increasing and where it is decreasing.

Differentiate:

f'(x) = 2x - 4.

Solve the inequality $f'(x) > 0$ :

2x - 4 > 0 \quad \Rightarrow \quad x > 2.

So $f$ is increasing for $x > 2$ . Solve $f'(x) < 0$ :

2x - 4 < 0 \quad \Rightarrow \quad x < 2.

So $f$ is decreasing for $x < 2$ . At $x = 2$ the derivative is zero, and the curve has a minimum.

2. A cubic with two turning points

When the derivative is a quadratic, the sign changes at each of its roots. A sign table keeps the intervals straight.

Example 2. Find where $f(x) = x^3 - 3x^2 + 1$ is increasing and where it is decreasing.

Differentiate and factorise:

f'(x) = 3x^2 - 6x = 3x(x - 2).

The roots are $x = 0$ and $x = 2$ . Test a value in each interval.

Interval	Test value	Sign of $f'(x)$	Behaviour of $f$
$x < 0$	$x = -1$	$3(-1)(-3) = 9 > 0$	increasing
$0 < x < 2$	$x = 1$	$3(1)(-1) = -3 < 0$	decreasing
$x > 2$	$x = 3$	$3(3)(1) = 9 > 0$	increasing

So $f$ is increasing for $x < 0$ and for $x > 2$ , and decreasing for $0 < x < 2$ . The curve rises, falls, then rises again.

Example 3. Find the intervals on which $f(x) = 2x^3 - 9x^2 + 12x$ is decreasing.

Differentiate and factorise:

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2).

The roots are $x = 1$ and $x = 2$ . Test $x = 0$ , $x = 1.5$ and $x = 3$ .

Interval	Test value	Sign of $f'(x)$	Behaviour of $f$
$x < 1$	$x = 0$	$6(-1)(-2) = 12 > 0$	increasing
$1 < x < 2$	$x = 1.5$	$6(0.5)(-0.5) = -1.5 < 0$	decreasing
$x > 2$	$x = 3$	$6(2)(1) = 12 > 0$	increasing

So the function is decreasing only on the interval $1 < x < 2$ .

Watch out. Three mistakes are common with increasing and decreasing functions.

Saying "increasing" because the values are positive. A function is increasing where its gradient is positive. The function $y = x^2 - 4$ is increasing for $x > 0$ , even though its values are negative until $x = 2$ .

Testing one point and assuming the rest. The sign of $f'$ can change at each root. Use a sign table with one test value in every interval.

Writing $\le$ and $\ge$ as open or closed carelessly. Strictly increasing and strictly decreasing describe intervals where the derivative is not zero. At a turning point itself, the gradient is zero. Re-check: after you list the intervals, confirm that the derivative changes sign at each boundary.

You should now be able to

decide where a function is increasing or decreasing from the sign of $f'$ ;
solve $f'(x) > 0$ and $f'(x) < 0$ for a polynomial;
use a sign table to handle more than one root of $f'$ ;
sketch a curve whose intervals of increase and decrease you have found.