Mathematics guide, Unit 5, step 18 of 33Previous step Next step

Step 18: Stationary Points, Concavity and Curve Sketching

By the EconTutorials team. 11-minute read with 6 practice questions.

SyllabusAP 5.6–5.9, 5.12 · 9709 P1 1.7 · IB SL 5.7–5.8

In 30 seconds

The second derivative says which way a graph bends. $ f” > 0 $: concave up, like a cup. $ f” < 0 $: concave down, like a cap. A point of inflection is where the concavity changes, which needs a sign change of $ f” $, not just $ f” = 0 $.
Second derivative test at a stationary point ($ f'(c) = 0 $): $ f”(c) < 0 $ means a local maximum, $ f”(c) > 0 $ a local minimum, and $ f”(c) = 0 $ means the test tells you nothing, so go back to the sign of $ f’ $.
On the graph of $ f’ $: zeros where $ f’ $ changes sign are maxima or minima of $ f $; turning points of $ f’ $ are inflections of $ f $.
Exam point: for an implicit curve with $ \dfrac{dy}{dx} = \dfrac{N}{D} $, tangents are horizontal where $ N = 0 $ (and $ D \neq 0 $) and vertical where $ D = 0 $ (and $ N \neq 0 $). Always solve together with the curve’s equation.

Video lesson coming soonA short animated lesson for this step will appear here.

Concavity and the second derivative

In Step 17 the first derivative told us which way a graph goes: up where $ f’ > 0 $, down where $ f’ < 0 $. Two curves can both be rising, though, and still look completely different. One gets steeper and steeper, like a rocket taking off; the other rises more and more gently, like a plane levelling out. The difference is in how the gradient itself is changing, and that is measured by the second derivative $ f”(x) $, the derivative of $ f'(x) $.

Key termConcave up on an interval: $ f’ $ is increasing there, so the graph bends upwards like a cup and lies above its tangents. This holds if $ f”(x) > 0 $ on the interval. Concave down is the reverse: $ f’ $ decreasing, $ f”(x) < 0 $, graph bending like a cap and lying below its tangents.

Key termPoint of inflection: a point on the graph where $ f $ is continuous and the concavity changes, from up to down or from down to up.

To find inflections, solve $ f”(x) = 0 $ (and note where $ f” $ does not exist), then check that $ f” $ really changes sign there. The check is not optional, as the common mistake later in this step shows.

Worked example

Find the intervals on which $ f(x) = xe^{-x} $ is concave up and concave down, and find its point of inflection.

Step 1Product rule: $ f'(x) = e^{-x}-xe^{-x} = (1-x)e^{-x} $.

Step 2Again: $ f”(x) = -e^{-x}-(1-x)e^{-x} = (x-2)e^{-x} $.

Step 3$ e^{-x} > 0 $ always, so the sign of $ f” $ is the sign of $ x-2 $. Concave down for $ x < 2 $, concave up for $ x > 2 $.

Step 4$ f” $ changes sign at $ x = 2 $, and $ f(2) = 2e^{-2} \approx 0.271 $.

The point of inflection is $ (2, 2e^{-2}) $. The maximum at $ x = 1 $ (Q5 of Step 17) sits in the concave-down part, exactly where a peak should be.

Concavity of y = xe⁻ˣ

y = xe⁻ˣ

Figure 1. Left of x = 2 the curve bends like a cap; right of it, like a cup, as it flattens towards the x-axis. The change happens at the point of inflection (2, 2e⁻²).

An economics example: diminishing returns. A firm’s total cost is $ C = Q^3-12Q^2 + 60Q + 50 $ dollars. Then $ MC = C’ = 3Q^2-24Q + 60 $ and $ C” = 6Q-24 $, which changes sign at $ Q = 4 $. Below 4 units, $ C $ is concave down: each extra unit costs less than the one before. Above 4 units it is concave up: costs rise ever faster. The inflection of total cost at $ Q = 4 $ is exactly the minimum of marginal cost, $ MC(4) = 48-96 + 60 = \$12 $. Economists call it the point of diminishing marginal returns.

The second derivative test

The sign of $ f” $ at a stationary point gives a quick way to classify it.

Key termSecond derivative test: suppose $ f'(c) = 0 $. If $ f”(c) < 0 $, $ f $ has a local maximum at $ c $. If $ f”(c) > 0 $, a local minimum. If $ f”(c) = 0 $, the test is inconclusive: use the first derivative test instead.

Why does it work? At $ c $ the gradient is zero. If $ f”(c) < 0 $, the gradient is decreasing as $ x $ passes through $ c $, so it goes from positive to negative: up, then down. That is precisely the first derivative test’s sign change for a maximum, read off from the rate at which the gradient changes.

Worked example

Find the coordinates of the stationary points of $ y = 2x^3 + 3x^2-12x + 4 $ and determine their nature.

Step 1$ \dfrac{dy}{dx} = 6x^2 + 6x-12 = 6(x + 2)(x-1) $, so $ x = -2 $ or $ x = 1 $.

Step 2$ y(-2) = -16 + 12 + 24 + 4 = 24 $ and $ y(1) = 2 + 3-12 + 4 = -3 $.

Step 3$ \dfrac{d^2y}{dx^2} = 12x + 6 $. At $ x = -2 $ it is $ -18 < 0 $; at $ x = 1 $ it is $ 18 > 0 $.

$ (-2, 24) $ is a maximum point and $ (1, -3) $ is a minimum point.

Exam tipCambridge 9709 Paper 1 (1.7) lets you use either the second derivative or the sign of $ \dfrac{dy}{dx} $ either side unless the question names a method, but you must write a conclusion: “$ \dfrac{d^2y}{dx^2} = 18 > 0 $, so $ (1, -3) $ is a minimum.” Note that 9709 Paper 1 does not examine points of inflexion; AP (5.6 to 5.7) and IB AA SL (5.7 to 5.8) do, including inflections with zero and non-zero gradient.

Inflections and sketching a curve

When $ f”(c) = 0 $ at a stationary point, anything can happen. Here is a curve with all three kinds of stationary point plus two inflections that are not stationary at all.

Worked example

Sketch $ y = 3x^5-5x^3 $, showing its stationary points and points of inflection.

Step 1$ y’ = 15x^4-15x^2 = 15x^2(x-1)(x + 1) $: stationary at $ x = -1, 0, 1 $. $ y” = 60x^3-30x = 30x(2x^2-1) $.

Step 2$ x = -1 $: $ y” = -30 < 0 $, a maximum at $ (-1, 2) $. $ x = 1 $: $ y” = 30 > 0 $, a minimum at $ (1, -2) $.

Step 3$ x = 0 $: $ y” = 0 $, so the test fails. Near 0, $ 15x^2 \ge 0 $ and $ (x-1)(x + 1) < 0 $, so $ y’ < 0 $ on both sides: no sign change. Since $ y” $ does change sign at 0, $ (0, 0) $ is a stationary point of inflection.

Step 4$ y” = 0 $ also at $ x = \pm\tfrac{1}{\sqrt{2}} $, with a sign change each time, but $ y’ \neq 0 $ there: non-stationary inflections at $ \left(\pm\tfrac{1}{\sqrt{2}}, \mp\tfrac{7\sqrt{2}}{8}\right) \approx (\pm 0.707, \mp 1.237) $.

Step 5Intercepts: $ y = x^3(3x^2-5) = 0 $ at $ x = 0 $ and $ x = \pm\sqrt{5/3} \approx \pm 1.29 $. End behaviour: $ y \to \infty $ as $ x \to \infty $, $ y \to -\infty $ as $ x \to -\infty $.

The curve is symmetric about the origin (an odd function), which is a useful check on Figure 2.

y = 3x⁵ − 5x³: stationary points and inflections

y = 3x⁵ − 5x³tangent at a non-stationary inflection

Figure 2. A maximum, a minimum, a stationary inflection at the origin (horizontal tangent) and two non-stationary inflections, where the gold tangents are steep but the concavity still changes.

Feature	How to find it
Intercepts	Put $ x = 0 $ for the $ y $-intercept; solve $ y = 0 $ for the roots
Stationary points	Solve $ f'(x) = 0 $; classify with $ f” $ or the sign of $ f’ $
Inflections	Solve $ f”(x) = 0 $ and check that $ f” $ changes sign
End behaviour	The leading term decides what happens as $ x \to \pm\infty $

Common mistake“$ f”(c) = 0 $, so there is an inflection at $ c $.” Not necessarily. For $ f(x) = x^4 $, $ f”(0) = 0 $, but $ f”(x) = 12x^2 \ge 0 $ on both sides, so the curve is concave up throughout and $ (0, 0) $ is a minimum. You need a sign change of $ f” $. On AP, “$ f” = 0 $” alone earns no credit for an inflection.

Reading f, f′ and f″ from one another

A favourite AP question gives you only the graph of $ f’ $ and asks about $ f $. Every feature of $ f $ appears on it, one level down.

On the graph of $ f’ $	What it means for $ f $
above / below the $ x $-axis	$ f $ increasing / decreasing
crosses the axis from $ + $ to $-$ (from $-$ to $ + $)	local maximum (minimum) of $ f $
touches the axis without crossing	horizontal tangent, no extremum
rising / falling	$ f $ concave up / concave down
a turning point of $ f’ $	a point of inflection of $ f $

Worked example

Figure 3 shows the graph of $ f'(x) = (x + 2)(x-1)^2 $. Using it, describe the behaviour of $ f $.

Step 1$ f’ < 0 $ for $ x < -2 $ and $ f’ > 0 $ for $ x > -2 $ (except at $ x = 1 $). So $ f $ decreases, then increases.

Step 2At $ x = -2 $, $ f’ $ crosses from $-$ to $ + $: a local minimum of $ f $. At $ x = 1 $, $ f’ $ touches zero without changing sign: a horizontal tangent but no extremum.

Step 3$ f’ $ rises until $ x = -1 $, falls until $ x = 1 $, then rises again. Check: $ f”(x) = 3(x-1)(x + 1) $. So $ f $ is concave up for $ x < -1 $, concave down for $ -1 < x < 1 $, concave up for $ x > 1 $.

$ f $ has inflections at $ x = -1 $ and $ x = 1 $; the one at $ x = 1 $ is stationary. The graph of $ f’ $ cannot tell you any heights of $ f $, such as $ f(0) $.

The graph of f′(x) = (x + 2)(x − 1)²

y = f′(x)

Figure 3. Read f from f′: below the axis f falls, above it f rises; where f′ turns, f changes concavity. At x = 1, f′ touches zero without changing sign, so f has a stationary inflection there.

Exam tipJustify concavity from a graph of $ f’ $ in words that name $ f’ $: “$ f $ is concave down on $ (-1, 1) $ because $ f’ $ is decreasing there.” Writing “because $ f’ $ is negative” is a different (and wrong) reason.

Behaviour of implicit relations

For a curve given implicitly, such as a tilted ellipse, implicit differentiation (Step 14) gives $ \dfrac{dy}{dx} $ as a fraction $ \dfrac{N(x, y)}{D(x, y)} $. The tangent is horizontal where $ N = 0 $ and $ D \neq 0 $, and vertical where $ D = 0 $ and $ N \neq 0 $. Each condition gives a relation between $ x $ and $ y $ that you solve together with the curve’s equation. This is AP topic 5.12. Cambridge 9709 Paper 3 also asks for points where the tangent to an implicit curve is parallel to an axis, but uses only the first derivative; the second-derivative part below is for AP (and IB HL, 5.14).

Worked example

For the curve $ x^2-xy + y^2 = 3 $, find the points where the tangent is horizontal or vertical, and decide whether each horizontal-tangent point is a local maximum or minimum.

Step 1Differentiate: $ 2x-y-x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0 $, so $ \dfrac{dy}{dx} = \dfrac{y-2x}{2y-x} $.

Step 2Horizontal: $ y = 2x $. Substitute: $ x^2-2x^2 + 4x^2 = 3x^2 = 3 $, so $ (1, 2) $ and $ (-1, -2) $. There $ 2y-x = \pm 3 \neq 0 $.

Step 3Vertical: $ x = 2y $. Substitute: $ 4y^2-2y^2 + y^2 = 3y^2 = 3 $, so $ (2, 1) $ and $ (-2, -1) $.

Step 4Differentiate $ \dfrac{dy}{dx} $ by the quotient rule. At a point where $ \dfrac{dy}{dx} = 0 $ and $ y-2x = 0 $, everything collapses to $ \dfrac{d^2y}{dx^2} = \dfrac{-2}{2y-x} $.

Step 5At $ (1, 2) $: $ -\tfrac{2}{3} < 0 $, a local maximum (the top of the curve). At $ (-1, -2) $: $ +\tfrac{2}{3} > 0 $, a local minimum (the bottom).

Figure 4 shows all four points on the tilted ellipse.

The curve x² − xy + y² = 3

x² − xy + y² = 3

Figure 4. Horizontal tangents (gold) where y = 2x and vertical tangents (orange) where x = 2y. The top of the curve is (1, 2) and the bottom is (−1, −2).

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find where $ f(x) = x^3-3x^2 + 4 $ is concave up and concave down, and find its point of inflection. Show worked answer

$ f'(x) = 3x^2-6x $, so $ f”(x) = 6x-6 $.
$ f” < 0 $ for $ x < 1 $: concave down. $ f” > 0 $ for $ x > 1 $: concave up.
$ f” $ changes sign at $ x = 1 $, and $ f(1) = 1-3 + 4 = 2 $.
Point of inflection: $ (1, 2) $.

EasyQ2. Use the second derivative test to classify the stationary point of $ y = x + \dfrac{4}{x} $ for $ x > 0 $. Show worked answer

$ \dfrac{dy}{dx} = 1-\dfrac{4}{x^2} = 0 $ gives $ x^2 = 4 $, so $ x = 2 $ (as $ x > 0 $).
$ y = 2 + 2 = 4 $.
$ \dfrac{d^2y}{dx^2} = \dfrac{8}{x^3} = 1 > 0 $ at $ x = 2 $.
So $ (2, 4) $ is a minimum point.

MediumQ3. Find the coordinates of the stationary points of $ y = x^3-3x^2-24x + 5 $ and determine their nature. Show worked answer

$ \dfrac{dy}{dx} = 3x^2-6x-24 = 3(x-4)(x + 2) $, so $ x = 4 $ or $ x = -2 $.
$ y(-2) = -8-12 + 48 + 5 = 33 $; $ y(4) = 64-48-96 + 5 = -75 $.
$ \dfrac{d^2y}{dx^2} = 6x-6 $: $ -18 < 0 $ at $ x = -2 $, $ 18 > 0 $ at $ x = 4 $.
Maximum $ (-2, 33) $; minimum $ (4, -75) $.

MediumQ4. For $ y = x^4-6x^2 + 8x $, find and classify the stationary points, and find any non-stationary points of inflection. Show worked answer

$ y’ = 4x^3-12x + 8 = 4(x-1)^2(x + 2) $: stationary at $ x = -2 $ and $ x = 1 $.
$ y” = 12x^2-12 = 12(x-1)(x + 1) $.
$ x = -2 $: $ y” = 36 > 0 $, minimum at $ (-2, -24) $.
$ x = 1 $: $ y” = 0 $, test fails. $ (x-1)^2 \ge 0 $ and $ x + 2 > 0 $ near 1, so $ y’ > 0 $ on both sides: no extremum. $ y” $ changes sign at 1, so $ (1, 3) $ is a stationary point of inflection.
$ y” $ also changes sign at $ x = -1 $, where $ y’ = 16 \neq 0 $: a non-stationary inflection at $ (-1, -13) $.

MediumQ5. A function $ f $ has derivative $ f'(x) = x^2(x-3) $. Find the $ x $-coordinates of any local extrema and points of inflection of $ f $, with reasons. Show worked answer

$ f’ = 0 $ at $ x = 0 $ and $ x = 3 $. $ x^2 \ge 0 $, so the sign of $ f’ $ is the sign of $ x-3 $.
At $ x = 3 $, $ f’ $ changes from negative to positive: local minimum.
At $ x = 0 $, $ f’ $ is negative on both sides: no extremum.
$ f”(x) = 3x^2-6x = 3x(x-2) $, which changes sign at $ x = 0 $ and $ x = 2 $.
Points of inflection at $ x = 0 $ and $ x = 2 $.

ExamQ6. A curve has equation $ x^2 + 3y^2-2xy = 8 $. (a) Show that $ \dfrac{dy}{dx} = \dfrac{y-x}{3y-x} $. [2] (b) Find the coordinates of the points on the curve where the tangent is horizontal. [3] (c) Find the coordinates of the points where the tangent is vertical. [2] (d) Determine whether the curve has a local maximum or a local minimum at the horizontal-tangent point with positive coordinates. [2] Show worked answer

(a) $ 2x + 6y\dfrac{dy}{dx}-2y-2x\dfrac{dy}{dx} = 0 $. [M1]
$ \dfrac{dy}{dx}(6y-2x) = 2y-2x $, so $ \dfrac{dy}{dx} = \dfrac{y-x}{3y-x} $. [A1]
(b) Horizontal: $ y = x $. [M1]
Substitute: $ x^2 + 3x^2-2x^2 = 2x^2 = 8 $, so $ x = \pm 2 $. [M1]
Points $ (2, 2) $ and $ (-2, -2) $; there $ 3y-x = \pm 4 \neq 0 $. [A1]
(c) Vertical: $ x = 3y $, so $ 9y^2 + 3y^2-6y^2 = 6y^2 = 8 $, $ y = \pm\tfrac{2}{\sqrt{3}} $. [M1]
Points $ \left(2\sqrt{3}, \tfrac{2}{\sqrt{3}}\right) $ and $ \left(-2\sqrt{3}, -\tfrac{2}{\sqrt{3}}\right) $. [A1]
(d) Quotient rule; where $ \dfrac{dy}{dx} = 0 $ and $ y = x $, $ \dfrac{d^2y}{dx^2} = \dfrac{-(3y-x)}{(3y-x)^2} = \dfrac{-1}{3y-x} $. [M1]
At $ (2, 2) $: $ -\tfrac{1}{4} < 0 $, so a local maximum. [A1]
Total: 9 marks.

Check yourself

1. $ f”(x) = (x-2)^2 $. Does $ f $ have a point of inflection at $ x = 2 $?

$ (x-2)^2 \ge 0 $ on both sides, so $ f $ is concave up on both sides of 2. No change of concavity means no inflection.

2. At $ x = c $, $ f'(c) = 0 $ and $ f”(c) = -5 $. What does $ f $ have at $ c $?

A stationary point where the graph is concave down is a peak: by the second derivative test, $ f”(c) < 0 $ gives a local maximum.

3. The graph of $ f’ $ has a local minimum at $ x = 3 $. What can you say about $ f $ at $ x = 3 $?

$ f’ $ falls then rises, so $ f” $ goes from negative to positive: $ f $ changes from concave down to concave up, an inflection.

Your score: 0 out of 3

The short version

$ f” > 0 $: concave up (cup); $ f” < 0 $: concave down (cap). An inflection needs a sign change of $ f” $, not just $ f” = 0 $.
Second derivative test at $ f'(c) = 0 $: $ f”(c) < 0 $ maximum, $ f”(c) > 0 $ minimum, $ f”(c) = 0 $ inconclusive, so use the sign of $ f’ $.
A stationary inflection has $ f’ = 0 $ with no sign change of $ f’ $; a non-stationary inflection has $ f’ \neq 0 $.
On a graph of $ f’ $: sign changes are extrema of $ f $, turning points are inflections of $ f $, and heights of $ f $ cannot be read.
Implicit curves: horizontal tangent where the numerator of $ \dfrac{dy}{dx} $ is 0, vertical where the denominator is 0; solve with the curve’s equation.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

On the graph of \( f’ \)	What it means for \( f \)
above / below the \( x \)-axis	\( f \) increasing / decreasing
crosses the axis from \( + \) to \(-\) (from \(-\) to \( + \))	local maximum (minimum) of \( f \)
touches the axis without crossing	horizontal tangent, no extremum
rising / falling	\( f \) concave up / concave down
a turning point of \( f’ \)	a point of inflection of \( f \)

Step 18: Stationary Points, Concavity and Curve Sketching

Concavity and the second derivative

The second derivative test

Inflections and sketching a curve

Reading f, f′ and f″ from one another

Behaviour of implicit relations

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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