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

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.

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

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

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

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).

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

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.