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Differentiation – Increasing and Decreasing Functions

📘 Notes

Increasing and Decreasing Functions — 9709 (Year 12)

We use the derivative \(f'(x)\) to describe where a function is increasing, decreasing, or stationary on an interval.

1. What do “increasing” and “decreasing” mean?

Think about how the graph moves as you go from left to right along the \(x\)-axis.

Graph of an increasing function
Increasing: as \(x\) increases, \(f(x)\) increases.
Graph of a decreasing function
Decreasing: as \(x\) increases, \(f(x)\) decreases.
Graph of a constant function
Constant: as \(x\) increases, \(f(x)\) stays the same.

2. Definitions (using the graph)

Let \(f\) be a function and consider an interval \(I\) on the \(x\)-axis.

  • \(f\) is increasing on \(I\) if for any \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) \le f(x_2)\).
  • \(f\) is strictly increasing on \(I\) if \(x_1 < x_2\) implies \(f(x_1) < f(x_2)\).
  • \(f\) is decreasing on \(I\) if \(x_1 < x_2\) implies \(f(x_1) \ge f(x_2)\).
  • \(f\) is strictly decreasing on \(I\) if \(x_1 < x_2\) implies \(f(x_1) > f(x_2)\).

For A Level work we mostly use the derivative to decide where a function is increasing or decreasing.

3. Connection with the derivative

Suppose \(f\) is differentiable on an interval. Then:

  • If \(f'(x) > 0\) for all \(x\) in an interval, then \(f\) is increasing on that interval.
  • If \(f'(x) < 0\) for all \(x\) in an interval, then \(f\) is decreasing on that interval.
  • If \(f'(x) = 0\) for all \(x\) in an interval, then \(f\) is constant on that interval.

Geometrically, \(f'(x)\) is the gradient of the tangent: positive gradient → graph slopes upwards, negative gradient → graph slopes downwards.

4. Stationary points and the sign of \(f'(x)\)

A point where \(f'(x) = 0\) is called a stationary point (the tangent is horizontal).

Graph with a local maximum
Local maximum: \(f'(x)\) changes from positive to negative.
Graph with a local minimum
Local minimum: \(f'(x)\) changes from negative to positive.

Using the sign of \(f'(x)\):

  • If \(f'(x)\) changes from \(+\) to \(-\) at \(x = a\), then \(f\) has a local maximum at \(x = a\).
  • If \(f'(x)\) changes from \(-\) to \(+\) at \(x = a\), then \(f\) has a local minimum at \(x = a\).
  • If the sign of \(f'(x)\) does not change, the stationary point is not a max or min (e.g. point of inflection).
5. Worked example

Question

Let \[ f(x) = x^{3} - 6x^{2} + 9x. \] (i) Find the stationary points of \(f\).
(ii) State the intervals on which \(f\) is increasing and decreasing.

Solution

Step 1: Differentiate.

\[ \begin{aligned} f(x) &= x^{3} - 6x^{2} + 9x, \\ f'(x) &= 3x^{2} - 12x + 9. \end{aligned} \]

Step 2: Find stationary points by solving \(f'(x) = 0\).

\[ \begin{aligned} 3x^{2} - 12x + 9 &= 0 \\ x^{2} - 4x + 3 &= 0 \quad (\text{divide by } 3) \\ (x - 1)(x - 3) &= 0. \end{aligned} \] So the stationary points occur at \(x = 1\) and \(x = 3\).

Find the corresponding \(y\)-values: \[ \begin{aligned} f(1) &= 1 - 6 + 9 = 4, \\ f(3) &= 27 - 54 + 27 = 0. \end{aligned} \] Stationary points: \((1, 4)\) and \((3, 0)\).

Step 3: Use the sign of \(f'(x)\) to find where \(f\) is increasing/decreasing.

We already have \(f'(x) = 3x^{2} - 12x + 9 = 3(x - 1)(x - 3)\).

Interval Test value Sign of \(f'(x)\) Behaviour of \(f\)
\((-\infty, 1)\) \(x = 0\) \(f'(0) = 3(0 - 1)(0 - 3) = 9 > 0\) Increasing
\((1, 3)\) \(x = 2\) \(f'(2) = 3(1)(-1) = -3 < 0\) Decreasing
\((3, \infty)\) \(x = 4\) \(f'(4) = 3(3)(1) = 9 > 0\) Increasing

So:

  • \(f\) is increasing on \((-\infty, 1)\) and \((3, \infty)\).
  • \(f\) is decreasing on \((1, 3)\).

At \(x = 1\), \(f'(x)\) changes from positive to negative, so \((1, 4)\) is a local maximum.
At \(x = 3\), \(f'(x)\) changes from negative to positive, so \((3, 0)\) is a local minimum.