Where does a roller coaster reach its highest peak? Where does a thrown ball pause at the top of its arc? These moments share a mathematical signature: the gradient is momentarily zero. In this lesson, you will learn how to find these special points — called stationary points — and how to determine whether they are summits, valleys, or flat terraces.
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Imagine throwing a ball straight up into the air. At the very top of its flight, the ball is momentarily stationary before it starts falling back down. If the ball's height is $h(t)$, what do you think is true about the derivative $h'(t)$ at that exact instant? And how might the second derivative $h''(t)$ help you confirm that this point is a maximum rather than a minimum?
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Wrong: Integration and differentiation produce identical results.
Right: Integration is the reverse of differentiation, but includes an arbitrary constant (+C) for indefinite integrals.
📚 Core Content
A stationary point is a point on a curve where the gradient is zero — the tangent line is horizontal. To find them:
For example, for $f(x) = x^3 - 3x^2$:
$$f'(x) = 3x^2 - 6x = 3x(x - 2) = 0$$
So $x = 0$ or $x = 2$. The stationary points are $(0, 0)$ and $(2, -4)$.
Once you have found a stationary point at $x = a$:
Examine the sign of $f'(x)$ just before and just after $x = a$:
A horizontal point of inflection is a stationary point where the curve changes concavity but does not turn around. The gradient is zero, but the function continues increasing (or decreasing) on both sides.
The classic example is $y = x^3$ at $(0, 0)$. Here:
🧮 Worked Examples
🧪 Activities
1 $f(x) = x^2 - 4x + 5$
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2 $y = 2x^3 - 3x^2$
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3 $f(x) = x^4 - 2x^2$
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4 $y = x^3 + 3x^2 + 3x$
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1 Explain the difference between a turning point and a horizontal point of inflection. Give an example of each.
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2 A student finds a stationary point where $f''(x) = 0$ and immediately concludes it is a point of inflection. Why might this be incorrect?
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Earlier you thought about a ball at the top of its arc.
At the highest point, the velocity $h'(t) = 0$ because the ball is momentarily stationary. The second derivative $h''(t)$ is negative because gravity is pulling the ball downward — this confirms the point is a local maximum (concave down). This is exactly how physicists and engineers use calculus to find peaks in trajectories, from sports science to rocket design.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. (a) Find the coordinates of the stationary points of $f(x) = x^3 - 6x^2 + 9x + 2$. 2 MARKS
(b) Determine the nature of each stationary point. 2 MARKS
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7. A farmer has 60 metres of fencing and wants to enclose a rectangular paddock against an existing straight wall. The side parallel to the wall requires no fencing. 4 MARKS
(a) Show that the area of the paddock is $A = x(60 - 2x)$, where $x$ is the width perpendicular to the wall.
(b) Find the dimensions that maximise the area, and state the maximum area.
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8. The curve $y = ax^3 + bx^2 + cx + d$ has a stationary point at $(1, 4)$ and a point of inflection at $(0, 2)$. Find the values of $a$, $b$, $c$, and $d$. 4 MARKS
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1. $f'(x) = 2x - 4 = 0 \Rightarrow x = 2$, $f(2) = 1$. $f''(x) = 2 > 0$, so $(2, 1)$ is a local minimum.
2. $y' = 6x^2 - 6x = 6x(x - 1) = 0 \Rightarrow x = 0$ or $x = 1$. $y(0) = 0$, $y(1) = -1$. $y'' = 12x - 6$, so $y''(0) = -6 < 0$ (local max at $(0, 0)$) and $y''(1) = 6 > 0$ (local min at $(1, -1)$).
3. $f'(x) = 4x^3 - 4x = 4x(x^2 - 1) = 0 \Rightarrow x = 0, \pm 1$. $f(0) = 0$, $f(\pm 1) = -1$. $f''(x) = 12x^2 - 4$, so $f''(0) = -4 < 0$ (local max at $(0, 0)$), $f''(\pm 1) = 8 > 0$ (local min at $(\pm 1, -1)$).
4. $y' = 3x^2 + 6x + 3 = 3(x + 1)^2 = 0 \Rightarrow x = -1$, $y(-1) = -1 + 3 - 3 = -1$. $y'' = 6x + 6$, so $y''(-1) = 0$ (inconclusive). First derivative test: $y' > 0$ on both sides, so $(-1, -1)$ is a horizontal point of inflection.
1. A turning point is a stationary point where the function changes direction (from increasing to decreasing or vice versa), such as $y = x^2$ at $(0, 0)$. A horizontal point of inflection is stationary but the function continues in the same direction on both sides, such as $y = x^3$ at $(0, 0)$.
2. $f''(x) = 0$ at a stationary point is inconclusive. The point could be a maximum, minimum, or horizontal inflection. The first derivative test must be used to determine the nature.
1. A — $f'(x) = 2x - 6 = 0 \\Rightarrow x = 3$.
2. B — $f'(x) = 3x^2 - 6x$, $f''(x) = 6x - 6$. $f''(2) = 6 > 0$, so it is a local minimum.
3. B — By definition, a stationary point is where the gradient $f'(x)$ equals zero.
4. C — $f'(x) = 3x^2$ is zero at $x = 0$, and $f'(x) > 0$ on both sides. This is a horizontal point of inflection.
5. C — $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0$, so $x = \\pm 1$. There are two stationary points.
6. B — $f''(a) > 0$ means the curve is concave up at $x = a$, which indicates a local minimum.
7. B — When $f'(x)$ changes from negative to positive, the function changes from decreasing to increasing, indicating a local minimum.
Q6 (4 marks): $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3) = 0$ [1]. Stationary points: $f(1) = 6 \Rightarrow (1,6)$ and $f(3) = 2 \Rightarrow (3,2)$ [1]. $f''(x) = 6x - 12$ [1]. $f''(1) = -6 < 0 \Rightarrow$ local maximum at $(1,6)$; $f''(3) = 6 > 0 \Rightarrow$ local minimum at $(3,2)$ [1].
Q7 (4 marks): Constraint $2x + y = 60 \Rightarrow y = 60 - 2x$, so $A = xy = x(60-2x)$ [1]. $\frac{dA}{dx} = 60 - 4x = 0 \Rightarrow x = 15$ [1]. Length $y = 60 - 2(15) = 30$ m [1]. Maximum area $= 15 \times 30 = 450$ m$^2$ [1].
Q8 (4 marks): $y(0) = 2 \Rightarrow d = 2$ [1]. Point of inflection at $(0, 2)$ means $y''(0) = 0$, and since $y'' = 6ax + 2b$, this gives $b = 0$ [1]. Stationary point at $(1, 4)$ gives $a + c + 2 = 4$ so $a + c = 2$, and $3a + c = 0$ [1]. Solving simultaneously yields $a = -1$ and $c = 3$ [1].
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