Course Notes Practice Past Paper Archive

Stationary Points & Curve Sketching0%

Differentiation · Topic 10 of 15

Stationary Points & Curve Sketching

Video lesson · from 55:373 worked examples

One lesson video covers all of Differentiation, so it opens at 55:37 for this topic — not from the beginning.

Theory

Points where the curve is neither increasing nor decreasing are called stationary points. At these points, the tangent is horizontal.

We can find stationary points by solving:

\frac{dy}{dx} = 0 \quad \text{or} \quad f'(x) = 0

There are four possible types of stationary point:

Minimum Turning Point

Maximum Turning Point

Rising Point of Inflection

Falling Point of Inflection

A stationary point's nature is determined by its behaviour to the left and right, using a nature table.

⚠️ Common Examiner Traps

The nature table has minimum requirements: a table that does not meet them scores nothing. It must show the $x$ values either side of the stationary point, the sign of $\frac{dy}{dx}$ in each region, and the shape of the slope — not just an answer.
Label it correctly: incorrect labelling within the table is a common and needless error. Say which $x$ values you are testing.
Evaluate carefully either side: numerical errors when evaluating the derivative near the point are common. Choose easy test values, and keep clear of the neighbouring stationary point.
Give the coordinates: solving $\frac{dy}{dx}=0$ gives only $x$ . Substitute back into the original function for $y$ and state the point.

Worked examples

Example 1

Find the stationary points of $y = 4x^3 - 2x^4$ and determine their nature.

Step 1: Differentiate and set to 0 for stationary points.

\begin{aligned} \frac{dy}{dx} &= 12x^2 - 8x^3 \\ 12x^2 - 8x^3 &= 0 \\ 4x^2(3 - 2x) &= 0 \end{aligned}

So $x = 0$ or $x = 1.5$ .

Step 2: Find corresponding $y$ coordinates.

At $x = 0$ , $y = 4(0)^3 - 2(0)^4 = 0$ . Point: (0, 0)
At $x = 1.5$ , $y = 4(1.5)^3 - 2(1.5)^4 = 13.5 - 10.125 = 3.375$ . Point: (1.5, 3.375)

Step 3: Nature table.

$x$	$\to 0$	0	$0 \to 1.5$	1.5	$1.5 \to$
$dy/dx$	+	0	+	0	-
Shape	/	—	/	—	\

(0,0) is a Rising Point of Inflection.

(1.5, 3.375) is a Maximum Turning Point.

Example 2

Find the stationary points of the curve with equation $y = x^3 - 3x^2 - 9x + 5$ and determine their nature.

Step 1: Differentiate and set to 0 to find stationary points.

\begin{aligned} \frac{dy}{dx} &= 3x^2 - 6x - 9 \\ 3x^2 - 6x - 9 &= 0 \\ 3(x^2 - 2x - 3) &= 0 \\ 3(x - 3)(x + 1) &= 0 \end{aligned}

So the stationary points occur at $x = 3$ and $x = -1$ .

Step 2: Find corresponding $y$ coordinates.

At $x = 3$ , $y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22$ . Point: (3, -22)
At $x = -1$ , $y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$ . Point: (-1, 10)

Step 3: Nature table.

$x$	$\to -1$	-1	$-1 \to 3$	3	$3 \to$
$dy/dx$	+	0	-	0	+
Shape	/	—	\	—	/

(-1, 10) is a Maximum Turning Point.

(3, -22) is a Minimum Turning Point.

Example 3

Sketch the curve with equation $y = x^3 - 3x^2$ . Show clearly the coordinates of the turning points and the points of intersection with the axes.

1. Intersections with axes:

y-axis: Let $x = 0$ .

y = 0^3 - 3(0)^2 = 0

So the point is (0, 0).

x-axis: Let $y = 0$ .

\begin{aligned} x^3 - 3x^2 &= 0 \\ x^2(x - 3) &= 0 \end{aligned}

So $x = 0$ or $x = 3$ . Points are (0, 0) and (3, 0).

2. Stationary points:

\frac{dy}{dx} = 3x^2 - 6x

Set to 0 for stationary points:

\begin{aligned} 3x^2 - 6x &= 0 \\ 3x(x - 2) &= 0 \end{aligned}

So $x = 0$ or $x = 2$ .

3. Coordinates and nature:

At $x = 0$ , $y = 0^3 - 3(0)^2 = 0$ . Point: (0, 0).
At $x = 2$ , $y = 2^3 - 3(2)^2 = 8 - 12 = -4$ . Point: (2, -4).

$x$	$\to 0$	0	$0 \to 2$	2	$2 \to$
$dy/dx$	+	0	-	0	+
Shape	/	—	\	—	/

(0, 0) is a Maximum Turning Point.

(2, -4) is a Minimum Turning Point.

4. Completed Diagram: