A retailer notices that as they raise prices, profit first rises, then peaks, then falls. Calculus tells us exactly where that peak occurs — and where the business is growing versus shrinking. In this lesson, you will learn how the sign of the derivative reveals when a function is increasing or decreasing, and how to locate and classify the critical turning points.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A company's monthly profit is modelled by $P(x) = -x^2 + 10x$, where $x$ is the advertising spend in thousands of dollars. Without calculating, predict: for what values of $x$ do you think profit is increasing? For what values is it decreasing? Where do you think the maximum profit occurs?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: If f'(x) = 0 at a point, that point must be either a maximum or a minimum.
Right: A stationary point can also be a horizontal point of inflection. You must test the sign of f'(x) on either side to classify it.
Core Content
If $f'(x) > 0$ on an interval, the function $f$ is increasing on that interval. If $f'(x) < 0$, the function is decreasing.
This is about the gradient of the curve, not the value of the function. A function can be increasing even when its values are negative, and decreasing even when its values are positive.
A stationary point occurs where the gradient is zero: $f'(x) = 0$.
To classify a stationary point, we use the first derivative test: examine the sign of $f'(x)$ just before and just after the stationary point.
Worked Examples
$f'(x) = 3x^2 - 6x = 3x(x-2)$.
Stationary points at $x = 0$ and $x = 2$.
$f'(-1) = 9 > 0$, $f'(1) = -3 < 0$, $f'(3) = 9 > 0$.
Increasing on $(-\infty, 0) \cup (2, \infty)$; decreasing on $(0, 2)$.
Activities
1 $f(x) = x^2 - 6x + 5$
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2 $y = x^3 - 3x^2$
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3 $f(x) = 2x^3 - 9x^2 + 12x$
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4 $y = x^4 - 2x^2$
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1 Explain why a function can be increasing at a point where $f''(x) < 0$. Use an example.
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2 A student finds $f'(2) = 0$ and immediately concludes $(2, f(2))$ is a local minimum. What step did they miss, and why does it matter?
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Earlier you thought about the profit model $P(x) = -x^2 + 10x$.
The derivative is $P'(x) = -2x + 10$. Profit is increasing when $P'(x) > 0$, which occurs when $x < 5$. It is decreasing when $x > 5$. The maximum profit occurs at $x = 5$, where $P'(5) = 0$ and the derivative changes from positive to negative — confirming a local maximum.
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 intervals where $f(x) = x^3 - 6x^2 + 9x + 2$ is increasing and where it is decreasing. (b) Find the coordinates and nature of the stationary points. 4 marks
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7. A company's profit is $P(x) = -2x^2 + 24x - 30$ for $x \ge 0$, where $x$ is price in dollars. (a) Find the interval where profit is increasing. (b) Find the price that maximises profit and the maximum profit. 4 marks
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8. Explain why the first derivative test always works for classifying stationary points of a continuous differentiable function, and describe a situation where it might be preferable to the second derivative test. 4 marks
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1. $f'(x) = 2x - 6$. Stationary point at $x = 3$, $f(3) = -4$. $f'(x) < 0$ for $x < 3$ and $f'(x) > 0$ for $x > 3$, so decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$. Local minimum at $(3, -4)$.
2. $f'(x) = 3x^2 - 6x = 3x(x-2)$. Stationary points at $x = 0$, $f(0)=0$ and $x = 2$, $f(2)=-4$. Increasing on $(-\infty, 0) \cup (2, \infty)$, decreasing on $(0, 2)$. Local max at $(0, 0)$, local min at $(2, -4)$.
3. $f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$. Stationary points at $x = 1$, $f(1)=5$ and $x = 2$, $f(2)=4$. Increasing on $(-\infty, 1) \cup (2, \infty)$, decreasing on $(1, 2)$. Local max at $(1, 5)$, local min at $(2, 4)$.
4. $f'(x) = 4x^3 - 4x = 4x(x-1)(x+1)$. Stationary points at $x = -1$, $f(-1)=-1$; $x = 0$, $f(0)=0$; $x = 1$, $f(1)=-1$. Increasing on $(-1, 0) \cup (1, \infty)$, decreasing on $(-\infty, -1) \cup (0, 1)$. Local min at $(-1, -1)$, local max at $(0, 0)$, local min at $(1, -1)$.
1. $f''(x)$ measures concavity, not direction. For example, $f(x) = x^2$ on $(0, 1)$ is increasing ($f'(x) = 2x > 0$) but concave up ($f''(x) = 2 > 0$). A better example: $f(x) = \ln x$ is always increasing ($f'(x) = 1/x > 0$) yet concave down ($f''(x) = -1/x^2 < 0$). Direction and concavity are independent properties.
2. The student missed the first derivative test (or second derivative test). They need to check the sign of $f'(x)$ on both sides of $x = 2$. Without this, $x = 2$ could be a local maximum, a local minimum, or a horizontal point of inflection.
1. B — $f'(x) > 0$ means the gradient is positive, so $f(x)$ is increasing.
2. B — A stationary point is defined by $f'(x) = 0$.
3. B — A sign change from negative to positive indicates the curve bottoms out, giving a local minimum.
4. B — $f'(x) = 3x^2 - 6x = 3x(x-2)$, which is negative on $(0, 2)$.
5. C — When the sign of $f'(x)$ does not change, the stationary point is a horizontal point of inflection.
6. B — $f'(x) = 2x - 4 > 0$ when $x > 2$.
7. B — $P'(x)$ changes from positive to negative at $x = 5$, indicating a local maximum. In context, profit is maximised.
Q6 (4 marks): (a) $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$ [1]. Increasing on $(-\infty, 1)$ and $(3, \infty)$; decreasing on $(1, 3)$ [1]. (b) Stationary points at $(1, 6)$ and $(3, 2)$ [1]. Local max at $(1, 6)$ and local min at $(3, 2)$ by first derivative test [1].
Q7 (4 marks): (a) $P'(x) = -4x + 24$. $P'(x) > 0$ when $x < 6$, so profit is increasing for $0 \le x < 6$ [2]. (b) Maximum at $x = 6$ [1]. $P(6) = -72 + 144 - 30 = 42$, so maximum profit is $42 [1].
Q8 (4 marks): The first derivative test works because the sign of $f'(x)$ tells us whether the function is rising or falling on each side of the stationary point [2]. It is preferable to the second derivative test when $f''(x) = 0$ at the stationary point (so the second derivative test is inconclusive) or when $f''(x)$ is difficult to compute [2].