Understanding Increasing and Decreasing Functions Revision Notes | GCSE AQA Higher Further Maths

1. What Does a Function Actually Do?

Picture walking along a hill. When you are climbing uphill, the ground rises beneath your feet — that is an increasing section. When you descend, the ground falls away — that is a decreasing section. Functions behave in exactly the same way.

The key to describing this behaviour mathematically is the gradient. A positive gradient means the function is going uphill (increasing); a negative gradient means it is going downhill (decreasing). That single idea unlocks everything in this topic.

The reverse holds for a decreasing function. When $\frac{dy}{dx} < 0$, the curve slopes downward from left to right, and $y$ falls as $x$ grows. Both descriptions refer to behaviour over an interval, not at a single isolated point.

The graph above shows $y = x^2 - 4x + 3$. Notice how the curve falls from left to right until it reaches its lowest point, then rises again. The falling section is where the function is decreasing; the rising section is where it is increasing. The turning point sits exactly where the gradient changes sign.

2. Gradient Sign: The Core Rule

At GCSE, the gradient of a curve at any point is found by differentiating — that is, finding $\frac{dy}{dx}$. Once you have the derivative, you substitute an $x$-value and check whether the result is positive or negative.

The stationary point (where $\frac{dy}{dx} = 0$) acts as a dividing line. To the left of a minimum, the function is decreasing; to the right, it is increasing. The opposite pattern holds around a maximum. Identifying these boundary $x$-values is how you write down the intervals precisely.

3. Finding Increasing and Decreasing Intervals — Worked Examples

Knowing the rule is one thing; applying it to an actual function is where the marks are earned. The method always follows the same four-step pattern: differentiate, set the derivative to zero to find the boundary, test the sign of the derivative on each side, then state the intervals.

Worked example: Identifying Increasing and Decreasing Intervals for a Quadratic

Find the values of $x$ for which $f(x) = x^2 - 6x + 8$ is (a) decreasing and (b) increasing.

1. $f'(x) = 2x - 6$ — Bring the power down and reduce the index by 1 for each term. The constant 8 disappears. [M1 — correct differentiation]
2. $2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = 3$ — The gradient is zero at $x = 3$. This is the stationary point — the boundary between increasing and decreasing behaviour. [M1]
3. Try $x = 0$: $f'(0) = 2(0) - 6 = -6 < 0$ — The gradient is negative for $x < 3$, so the function is decreasing on this interval.
4. Try $x = 5$: $f'(5) = 2(5) - 6 = 10 - 6 = 4 > 0$ — The gradient is positive for $x > 3$, so the function is increasing on this interval.
5. Decreasing: $x < 3$ Increasing: $x > 3$ — Write the answer using inequality notation. The stationary point $x = 3$ is the boundary and is not included in either interval. [A1 for each correct interval]

Answer: $f(x)$ is decreasing for $x < 3$ and increasing for $x > 3$.

Notice that the method works regardless of the function's complexity — differentiate, find where the derivative equals zero, then check the sign on each side. The same approach applies to cubics and beyond.

Worked example: Increasing and Decreasing Intervals for a Cubic

Find the intervals on which $g(x) = x^3 - 3x^2 - 9x + 5$ is increasing.

1. $g'(x) = 3x^2 - 6x - 9$ — Apply the power rule to each term. [M1]
2. $3x^2 - 6x - 9 = 0$ $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$ $x = 3 \text{ or } x = -1$ — Divide through by 3 first to simplify, then factorise. There are two stationary points, so three intervals to check. [M1 for solving]
3. Try $x = -2$: $g'(-2) = 3(-2)^2 - 6(-2) - 9 = 3 \times 4 + 12 - 9 = 12 + 12 - 9 = 15 > 0$ — Gradient is positive, so $g(x)$ is increasing for $x < -1$.
4. Try $x = 0$: $g'(0) = 3(0)^2 - 6(0) - 9 = -9 < 0$ — Gradient is negative, so $g(x)$ is decreasing between the two stationary points.
5. Try $x = 4$: $g'(4) = 3(4)^2 - 6(4) - 9 = 3 \times 16 - 24 - 9 = 48 - 24 - 9 = 15 > 0$ — Gradient is positive, so $g(x)$ is increasing for $x > 3$.
6. Increasing: $x < -1$ and $x > 3$ — Both intervals where $g'(x) > 0$ are the answer. [A1 for both intervals stated correctly]

Answer: $g(x)$ is increasing for $x < -1$ and for $x > 3$.

4. Describing Behaviour at a Specific Point

Sometimes an exam question asks whether a function is increasing or decreasing at a specific value of $x$, rather than over an interval. The method is even more direct: substitute the $x$-value into $\frac{dy}{dx}$ and check the sign of the result.

Worked example: Checking Whether a Function is Increasing or Decreasing at a Point

Is $f(x) = 2x^3 - 3x^2 - 12x + 1$ increasing or decreasing at $x = -1$?

1. $f'(x) = 6x^2 - 6x - 12$ — Apply the power rule to each term. [M1]
2. $f'(-1) = 6(-1)^2 - 6(-1) - 12$ $= 6 \times 1 + 6 - 12$ $= 6 + 6 - 12$ $= 0$ — Note: $(-1)^2 = (-1) \times (-1) = 1$. The gradient equals zero at $x = -1$, which means this is a stationary point.
3. $f'(-1) = 0$ — A gradient of zero means the function is neither increasing nor decreasing at $x = -1$ — it is at a stationary point. [A1 for correct conclusion]

Answer: $f(x)$ is stationary (neither increasing nor decreasing) at $x = -1$, since $f'(-1) = 0$.

5. Practice: Apply the Method

The best way to consolidate this method is to work through problems independently. Each question below follows the same four-step structure you have seen in the worked examples — differentiate, find the boundary, test the sign, state the interval.

6. Exam Technique and Common Mistakes

This topic rewards students who are precise with their language and careful with their signs. The worked method is straightforward, but there are several predictable places where marks are dropped — and knowing them in advance puts you ahead.

Bringing it all together: the gradient sign is the single deciding factor for whether a function is increasing or decreasing. A positive $\frac{dy}{dx}$ means increasing; a negative $\frac{dy}{dx}$ means decreasing. Find the boundary by setting the derivative to zero, test the sign on each side, and state your intervals with strict inequalities — that is the complete, mark-winning method for this specification requirement.