Before you can find a stationary point you have to know the derivative exists. This lesson defines differentiability at a point, proves that it forces continuity, and shows the three shapes a curve can take where it is continuous and still has no derivative.
Commit to a prediction about unbroken curves.
A curve can be drawn from end to end without lifting your pen. Without looking ahead decide whether that guarantees it has a derivative at every point, and try to sketch a counterexample if you think it does not.
State continuity precisely and name the ways it fails.
Continuity · no break in the curve
We just saw that "draw it without lifting the pen" is the informal picture. That raises a question: an informal picture cannot settle a borderline case, so what is the precise condition? This card answers it → three things must hold at once, and a function fails to be continuous when any one of them fails.
A function $f$ is continuous at $x = a$ when all three of these hold:
Each of the three failures has a name and a look:
Quick check: For $f(x) = \dfrac{x^2 - 9}{x - 3}$, what happens at $x = 3$?
Define the derivative at a point and read it as a tangent.
We just saw what it takes for a curve to be unbroken at a point. That raises a question: unbroken says nothing about the direction the curve is heading, so what extra condition gives a gradient? This card answers it → a single finite limiting value for the difference quotient, which geometrically is a single non-vertical tangent.
$f$ is differentiable at $x = a$ exactly when that limit exists and is finite. Geometrically: there is one non-vertical tangent to the curve at the point, and zooming in on the point makes the curve look more and more like that straight line.
The limit is two-sided, so it exists only when the left-hand and right-hand difference quotients agree:
That two-sided requirement is where every counterexample in this lesson comes from. "Non-vertical" matters too: a vertical tangent means the limit is infinite, so the derivative does not exist even though the curve is perfectly smooth to look at.
$f$ is differentiable at $a$ when $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ exists and is FINITE, equivalently when the curve has one non-vertical tangent there.; The limit is two-sided: the left and right gradients must agree.
Pause, copy the limit definition of $f'(a)$, and the two conditions it hides (the one-sided limits must agree, and the value must be finite), into your book.
The implication runs one way only, and knowing which way is worth a mark on its own.
Why it holds. If $f'(a)$ exists then
$\displaystyle\lim_{h \to 0}\big(f(a+h) - f(a)\big) = \lim_{h \to 0} \left(\frac{f(a+h) - f(a)}{h} \times h\right) = f'(a) \times 0 = 0$
so $\displaystyle\lim_{h \to 0} f(a+h) = f(a)$, which is exactly continuity at $a$. The step needs $f'(a)$ to be finite; that is where the hypothesis is used.
Why the converse fails. $y = |x|$ is continuous at 0 and has no derivative there. So:
True or false: If a function is continuous at $x = a$, then it must be differentiable at $x = a$.
Recognise corners, cusps and vertical tangents.
Continuous but not differentiable · three ways
We just saw that continuity does not deliver a derivative. That raises a question: the syllabus asks you to identify these points from either the equation or the graph, so what exactly are you looking for? This card answers it → three recognisable shapes, each with its own signature in $f'$.
All three curves are continuous at the marked point, and none of them has a derivative there. Only the mechanism differs.
| shape | example | what the one-sided gradients do |
|---|---|---|
| Corner | $y = |x|$ at $x = 0$ | Both finite, but different: $-1$ and $+1$. |
| Cusp | $y = x^{2/3}$ at $x = 0$ | Both infinite, with opposite signs: $-\infty$ and $+\infty$. |
| Vertical tangent | $y = x^{1/3}$ at $x = 0$ | Both infinite with the same sign, so the tangent is vertical. |
For the last two, differentiating confirms it. $y = x^{1/3}$ gives $\dfrac{dy}{dx} = \dfrac{1}{3}x^{-2/3} = \dfrac{1}{3\sqrt[3]{x^2}}$, which grows without bound as $x \to 0$ from either side. $y = x^{2/3}$ gives $\dfrac{dy}{dx} = \dfrac{2}{3}x^{-1/3}$, which is large and negative on the left, large and positive on the right.
Continuous but not differentiable in three ways: CORNER (one-sided gradients finite but unequal), CUSP (infinite, opposite signs), VERTICAL TANGENT (infinite, same sign).; Also not differentiable wherever $f$ is discontinuous, by the contrapositive.
Pause, copy the three failure shapes with their standard examples $|x|$, $x^{2/3}$ and $x^{1/3}$, and what the one-sided gradients do in each, into your book.
Odd one out: Three of these functions are continuous at $x = 0$ but not differentiable there. Which one is differentiable at $x = 0$?
Test piecewise functions at their join.
Worked examples · 3 in a row, reveal as you go
$f(x) = x^2$ for $x \leq 1$ and $f(x) = 2x - 1$ for $x > 1$. Is $f$ continuous at $x = 1$? Is it differentiable there?
$g(x) = x^2$ for $x \leq 1$ and $g(x) = x$ for $x > 1$. Determine continuity and differentiability at $x = 1$.
$h(x) = x^2$ for $x \leq 2$ and $h(x) = ax + b$ for $x > 2$. Find $a$ and $b$ so that $h$ is differentiable at $x = 2$.
Fill the blanks: $f(x) = x^2$ for $x \leq 3$ and $f(x) = mx + c$ for $x > 3$, and $f$ is differentiable at $x = 3$.
$m =$
$c =$
$f'(3) =$
Read the traps, work the quick-fire set, then revisit.
Common errors · traps that cost marks
Quick-fire practice · classify and test
State the three conditions for $f$ to be continuous at $x = a$, then name the discontinuity in $y = \dfrac{1}{x - 2}$ at $x = 2$.
$f(x) = |x - 4|$. Is $f$ differentiable at $x = 4$? Give the one-sided gradients.
$f(x) = 3x + 1$ for $x \leq 2$ and $f(x) = x^2 + k$ for $x > 2$. Find $k$ making $f$ continuous, then say whether it is differentiable at $x = 2$.
Explain why a function with a jump discontinuity at $x = a$ can never be differentiable there, without computing any limit.
Sketch a single curve that is continuous everywhere but fails to be differentiable at exactly two points, and label what happens at each.
Match each function to what happens at $x = 0$:
Earlier you were asked whether a curve you can draw without lifting your pen must have a derivative everywhere. It need not. Drawing without lifting is continuity; having a derivative is smoothness, and smoothness is strictly stronger. There are three counterexamples to reach for: a corner such as $y = |x|$, where the one-sided gradients are finite but unequal; a cusp such as $y = x^{2/3}$, where they are infinite with opposite signs; and a vertical tangent such as $y = x^{1/3}$, where they are infinite with the same sign. The implication only ever runs differentiable $\implies$ continuous.
Everything that follows in this focus area assumes the derivative exists where you use it. Stationary points are found by solving $f'(x) = 0$, which silently supposes $f'$ is defined; a corner can be a genuine maximum with no stationary point at all. Keeping that in mind is what stops a curve-sketching answer from missing a turning point entirely.
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 $f(x) = \begin{cases} x^2 + 1 & x \leq 2 \\ 4x - 3 & x > 2 \end{cases}$
(a) Show that $f$ is continuous at $x = 2$. (b) Determine whether $f$ is differentiable at $x = 2$, showing both one-sided gradients. (c) State the value of $f'(2)$ if it exists, or explain precisely why it does not. (3 marks)
Q2. A function is defined by $g(x) = \begin{cases} ax^2 + 3 & x \leq 1 \\ bx + 1 & x > 1 \end{cases}$
(a) Write down the two equations that must hold for $g$ to be differentiable at $x = 1$, saying which condition each comes from. (b) Solve them for $a$ and $b$. (c) Explain why solving only one of the two equations cannot determine both constants. (3 marks)
Q3. A student writes: "To find all the turning points of a function, solve $f'(x) = 0$. Any solution is a turning point, and every turning point is a solution."
(a) Give a function and a point showing that a solution of $f'(x) = 0$ need not be a turning point. (b) Give a function and a point showing that a turning point need not be a solution of $f'(x) = 0$. (c) State a corrected version of the student's rule that survives both counterexamples. (3 marks)
Drill 1: (i) $f(a)$ is defined; (ii) $\lim_{x \to a} f(x)$ exists, so the one-sided limits agree; (iii) the limit equals $f(a)$. At $x = 2$, $y = \tfrac{1}{x-2}$ has an infinite discontinuity: $f(2)$ is undefined and the function grows without bound on both sides, so condition (i) fails and there is a vertical asymptote.
Drill 2: $f(x) = |x-4|$ equals $4 - x$ for $x < 4$ and $x - 4$ for $x > 4$. The left gradient is $-1$ and the right gradient is $+1$. These are finite but unequal, so $f$ is not differentiable at $x = 4$: there is a corner at $(4, 0)$. It is continuous there, since both pieces give 0.
Drill 3: Continuity requires $3(2) + 1 = 2^2 + k$, so $7 = 4 + k$ and $k = 3$. Gradients: the left piece has constant gradient 3, the right piece has gradient $2x = 4$ at $x = 2$. Since $3 \neq 4$, $f$ is continuous but not differentiable at $x = 2$ — a corner. Choosing $k$ can fix continuity but cannot fix the gradient mismatch.
Drill 4: Differentiable at $a$ implies continuous at $a$. Taking the contrapositive, not continuous at $a$ implies not differentiable at $a$. A jump discontinuity means $f$ is not continuous, so no derivative can exist there, and no limit needs computing.
Drill 5: One good answer: $y = |x| + |x - 3|$, which has corners at $x = 0$ and $x = 3$ and is continuous everywhere. Another: a curve with a corner at one point and a vertical tangent at another, for instance built from $|x|$ near 0 and $(x-5)^{1/3}$ near 5. Both are continuous throughout because neither piece jumps.
Q1 (3 marks): (a) $f(2) = 2^2 + 1 = 5$ from the first piece, which owns $x = 2$. The right-hand limit is $\lim_{x \to 2^+}(4x - 3) = 5$. Both equal 5, and each piece is continuous in its own right, so $f$ is continuous at $x = 2$ ✓ [1]. (b) Left gradient: $\tfrac{d}{dx}(x^2 + 1) = 2x$, which is 4 at $x = 2$. Right gradient: $\tfrac{d}{dx}(4x - 3) = 4$. Both equal 4 [1]. (c) The one-sided gradients agree and $f$ is continuous, so $f$ IS differentiable at $x = 2$ with $f'(2) = 4$. (The line $4x - 3$ is the tangent to $y = x^2 + 1$ at $x = 2$, which is why the two pieces join invisibly.) [1]
Q2 (3 marks): (a) From continuity, the two pieces must agree at $x = 1$: $a(1)^2 + 3 = b(1) + 1$, that is $a + 3 = b + 1$. From matching gradients, $2ax$ at $x = 1$ must equal $b$: $2a = b$ [1]. (b) Substituting $b = 2a$ into $a + 3 = b + 1$ gives $a + 3 = 2a + 1$, so $a = 2$ and $b = 4$. Check: both pieces give 5 at $x = 1$, and both gradients give 4 ✓ [1]. (c) Each equation alone is one linear equation in two unknowns, so it has infinitely many solutions — a whole line of $(a, b)$ pairs. Differentiability is genuinely two conditions, not one: matching gradients alone allows the pieces to be parallel but separated (a jump), and matching values alone allows them to meet at a corner. Only together do they pin down a unique pair [1].
Q3 (3 marks): (a) $f(x) = x^3$ at $x = 0$: $f'(x) = 3x^2$ so $f'(0) = 0$, but the curve is increasing on both sides. It is a horizontal point of inflection, not a turning point [1]. (b) $f(x) = |x|$ at $x = 0$: the function decreases then increases, so $(0, 0)$ is a genuine minimum, yet $f'(0)$ does not exist, so it is not a solution of $f'(x) = 0$ at all [1]. (c) A corrected rule: the turning points of $f$ occur among the points where $f'(x) = 0$ together with the points where $f'$ fails to exist; each candidate must then be tested, by the sign of $f'$ either side or by the second derivative, to decide whether it is a maximum, a minimum or neither. The two counterexamples show that the original rule is wrong in both directions: it admits points that are not turning points, and it misses turning points that exist [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 →