Differentiate twice and you get the gradient function of the gradient function. It tells you which way the curve bends, where that bending reverses, and — as a bonus — a faster way to classify most stationary points.
Commit to a prediction about rising but slowing.
A quantity is rising, but rising more and more slowly. Without looking ahead say what that tells you about the signs of $f'$ and $f''$, and sketch the shape of the curve.
Differentiate twice and read the concavity off the sign.
Repeated differentiation · the gradient of the gradient
We just saw that the sign of $f'$ classifies stationary points but says nothing about the shape between them. That raises a question: $f'$ is itself a function, so what does its derivative tell us? This card answers it → how fast the gradient is changing, which is exactly the curve's bending.
Differentiating twice is written in three equivalent ways, and all three appear in HSC papers:
$f''$ is the gradient function of $f'$. So $f'' > 0$ means the gradient is increasing, and $f'' < 0$ means the gradient is decreasing — regardless of whether the curve itself is going up or down.
Back to the hook. Rising but slowing means $f' > 0$ and $f'' < 0$: increasing and concave down. All four sign combinations occur, and being able to name them is worth a mark on its own.
$f''$ is the gradient function of $f'$: $f'' > 0$ is concave up, $f'' < 0$ is concave down.; Concavity is independent of whether the curve rises or falls, so all four sign combinations of $(f', f'')$ are possible.
Pause, copy the three notations for the second derivative, the two concavity rules, and the note that concavity is independent of direction, into your book.
Quick check: At a point a curve has $f'(x) < 0$ and $f''(x) > 0$. The curve is:
Locate inflections, and test rather than assume.
We just saw that the sign of $f''$ names the concavity. That raises a question: a curve can bend one way then the other, so what happens at the changeover? This card answers it → that is a point of inflection, and it is defined by the change, not by $f''$ being zero.
Since $f''$ must pass from one sign to the other, at an inflection $f''$ is zero (or undefined). That gives a search method, not a test:
Horizontal versus non-horizontal. An inflection where $f'$ also happens to be zero is a horizontal point of inflection, like the origin on $y = x^3$. Where $f' \neq 0$ the inflection is oblique — the curve is still climbing or falling as it changes its bend. Both are points of inflection; only the first is also a stationary point.
An inflection is where CONCAVITY CHANGES. Solve $f''(x) = 0$ to find candidates, then test the sign of $f''$ either side and keep only those that change.; $f''(c) = 0$ alone is not enough: $y = x^4$ has $f''(0) = 0$ and no inflection.
Pause, copy the definition of a point of inflection, the four-step search-and-test method, and the $y = x^4$ counterexample, into your book.
True or false: If $f''(c) = 0$ then the curve has a point of inflection at $x = c$.
Classify stationary points faster, and know when it fails.
We just saw concavity read from the sign of $f''$. That raises a question: at a stationary point the curve is level, so the concavity there decides whether it is a peak or a trough — can we use that? This card answers it → yes, and it replaces the whole sign table with one substitution, except in one case.
Suppose $f'(c) = 0$. Then:
The reasoning is short: level and concave down means the curve falls away on both sides, so a peak; level and concave up means it rises away on both sides, so a trough.
| at a stationary point $c$ | conclusion |
|---|---|
| $f''(c) < 0$ | local maximum |
| $f''(c) > 0$ | local minimum |
| $f''(c) = 0$ | inconclusive — use the first-derivative sign test |
At a stationary point: $f''(c) < 0$ gives a local max, $f''(c) > 0$ a local min.; $f''(c) = 0$ is INCONCLUSIVE, not an inflection. Fall back to the sign of $f'$, and say that you are doing so.
Pause, copy the second-derivative test in all three cases, including that $f''(c) = 0$ is inconclusive, and the three functions $x^4$, $-x^4$ and $x^3$ that show why, into your book.
Odd one out: Three of these statements are true. Which one is false?
Work three examples end to end.
Worked examples · 3 in a row, reveal as you go
For $f(x) = x^3 - 6x^2 + 9x$, find and classify the stationary points using the second derivative, and find the point of inflection.
Classify the stationary point of $f(x) = x^4$ at $x = 0$, and state whether the curve has a point of inflection there.
Find the points of inflection of $f(x) = x^4 - 6x^2 + 5$, and state the concavity on each interval.
Fill the blanks: $f(x) = x^3 - 6x^2 + 9x$.
$f''(1) =$
inflection at $x =$
the $y$-value there $=$
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · differentiate twice, test, classify
$f(x) = x^3 - 3x^2 + 2$. Find $f''(x)$, then classify both stationary points using the second-derivative test.
Find the point of inflection of $f(x) = x^3 - 3x^2 + 2$, showing the sign change.
State the concavity of $f(x) = x^4 - 6x^2 + 5$ on each of the three intervals determined by its inflections.
Give a function that is decreasing and concave up at $x = 1$, and justify both claims with derivatives.
$y = x^4$, $y = -x^4$ and $y = x^3$ all have $f'(0) = f''(0) = 0$. Classify the origin for each, and say what that shows about the second-derivative test.
Match each pair of derivative signs to the shape of the curve:
Earlier you were asked about a quantity rising ever more slowly. That is $f' > 0$ and $f'' < 0$: increasing, because the quantity is still going up, and concave down, because the rate of increase is falling. The curve rises and flattens. If $f''$ later becomes positive the curve has passed a point of inflection — the moment the slowdown stopped getting worse — which is a genuinely different event from the peak, where $f'$ itself reaches zero. Reporting an inflection as a peak is the classic misreading of exactly this shape.
You can now find turning points, classify them two ways, and describe the bending in between. The next lesson assembles all of it into a single sketching procedure — domain, intercepts, symmetry, stationary points, concavity, inflections and asymptotic behaviour — and then runs the process in reverse: reading the shape of $f$ from a graph of $f'$, and of $f'$ from a graph of $f$.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 1 for the checkpoint question bank.
Q1. Consider $f(x) = 2x^3 - 3x^2 - 12x + 5$.
(a) Find $f'(x)$ and $f''(x)$. (b) Find the stationary points and classify each using the second-derivative test. (c) Find the point of inflection, showing that the concavity genuinely changes there. (3 marks)
Q2. A curve has $f''(x) = 6(x - 2)(x + 4)$.
(a) State the intervals on which the curve is concave up and concave down. (b) State the $x$-coordinates of all points of inflection, justifying each. (c) You are also told $f'(2) = 0$. Classify the stationary point at $x = 2$, and explain why your answer to (b) is unaffected. (3 marks)
Q3. A student writes: "To find the points of inflection, solve $f''(x) = 0$. Every solution is a point of inflection, and at every point of inflection $f''$ is zero."
(a) Give a counterexample to the first half of the claim. (b) The second half is true for every function in this course. Explain why, and state the condition under which it could fail. (c) A classmate replies: "Then the two halves are the same statement." Evaluate that reply. (3 marks)
Drill 1: $f'(x) = 3x^2 - 6x = 3x(x - 2)$, so stationary at $x = 0$ and $x = 2$. $f''(x) = 6x - 6$. Then $f''(0) = -6 < 0$, a local maximum; $f''(2) = 6 > 0$, a local minimum. ($f(0) = 2$ and $f(2) = 8 - 12 + 2 = -2$.)
Drill 2: $f''(x) = 6x - 6 = 0$ gives $x = 1$. $f''(0) = -6 < 0$ and $f''(2) = 6 > 0$, so the sign genuinely changes and $x = 1$ is a point of inflection. $f(1) = 1 - 3 + 2 = 0$, so the point is $(1, 0)$.
Drill 3: $f''(x) = 12x^2 - 12 = 12(x-1)(x+1)$. For $x < -1$, both factors negative, so $f'' > 0$: concave up. For $-1 < x < 1$, $f'' < 0$: concave down. For $x > 1$, $f'' > 0$: concave up.
Drill 4: One good answer is $f(x) = e^{-x}$. Then $f'(x) = -e^{-x}$, so $f'(1) = -e^{-1} < 0$ and the curve is decreasing; and $f''(x) = e^{-x}$, so $f''(1) = e^{-1} > 0$ and it is concave up. ($f(x) = \tfrac{1}{x}$ on $x > 0$ works equally well.)
Drill 5: $y = x^4$: local minimum (the first-derivative sign runs $-$ then $+$). $y = -x^4$: local maximum ($+$ then $-$). $y = x^3$: horizontal point of inflection ($+$ then $+$). All three have $f'(0) = f''(0) = 0$, which shows that when $f''(c) = 0$ the second-derivative test cannot distinguish the cases at all — it is genuinely inconclusive, and the first-derivative sign test must be used instead.
Q1 (3 marks): (a) $f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$ and $f''(x) = 12x - 6$ [1]. (b) $f'(x) = 0$ at $x = 2$ and $x = -1$. $f''(2) = 18 > 0$, so $(2, -15)$ is a local minimum; $f''(-1) = -18 < 0$, so $(-1, 12)$ is a local maximum. (Values: $f(2) = 16 - 12 - 24 + 5 = -15$; $f(-1) = -2 - 3 + 12 + 5 = 12$.) [1]. (c) $f''(x) = 12x - 6 = 0$ gives $x = \tfrac{1}{2}$. Testing: $f''(0) = -6 < 0$ and $f''(1) = 6 > 0$, so the concavity changes from down to up and this is a genuine point of inflection. $f\!\left(\tfrac12\right) = \tfrac14 - \tfrac34 - 6 + 5 = -\tfrac32$, so the point is $\left(\tfrac{1}{2}, -\tfrac{3}{2}\right)$ [1].
Q2 (3 marks): (a) $f''(x) = 6(x-2)(x+4)$ is a positive-leading quadratic with roots $-4$ and $2$, so it is positive outside the roots and negative between them. Concave up for $x < -4$ and for $x > 2$; concave down for $-4 < x < 2$ [1]. (b) The candidates are $x = -4$ and $x = 2$. Both are simple roots, so $f''$ changes sign at each: at $x = -4$ from positive to negative, at $x = 2$ from negative to positive. Both are therefore genuine points of inflection [1]. (c) $f''(2) = 0$, so the second-derivative test is inconclusive at $x = 2$ — but here we have more information: $f''$ changes from negative to positive across $x = 2$, so $f'$ has a minimum there, and since that minimum value is $f'(2) = 0$, $f'$ is positive on both sides. The curve is therefore increasing either side of a level moment, so $x = 2$ is a horizontal point of inflection. Answer (b) is unaffected because an inflection is defined by the concavity changing, which is a statement about $f''$ alone; whether $f'$ also happens to vanish there only decides whether the inflection is horizontal or oblique [1].
Q3 (3 marks): (a) $f(x) = x^4$ has $f''(x) = 12x^2$, so $f''(0) = 0$, yet $f''$ is positive on both sides: the curve is concave up throughout and there is no inflection at the origin. So a solution of $f''(x) = 0$ need not be an inflection [1]. (b) At an inflection the concavity changes, so $f''$ takes opposite signs on the two sides. For every function in this course $f''$ is continuous wherever it is defined, and a continuous function cannot change sign without passing through zero, so $f'' = 0$ at the inflection. It could fail only if $f''$ were discontinuous there — for instance if $f''$ jumped straight from negative to positive, or did not exist at the point at all, as happens at a cusp [1]. (c) The reply is wrong, and the error is a common one: the two halves are converse statements, not the same statement. "Every solution of $f'' = 0$ is an inflection" and "every inflection solves $f'' = 0$" have opposite directions, and part (a) shows precisely that one can be false while the other is true. Together they would say the two sets coincide; in fact the inflections are a strict subset of the solutions of $f'' = 0$, which is exactly why the sign test in step 2 of the method cannot be skipped [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →