IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Continuity & Differentiability

Continuity at a point is the formal version of “the graph doesn’t jump there”. Differentiability is stronger: it requires continuity AND smoothness (no corner). The logical chain only goes one way — differentiable ⟹ continuous, but the converse fails at every corner. The famous counter-example is |x| at x = 0: continuous, but with a corner that kills differentiability.

📘 What you need to know

Continuity at x = a: ALL THREE conditions must hold — f(a) is defined; the two-sided limit lim_{x → a} f(x) exists; AND that limit equals f(a).
Two failure modes: f(a) is undefined (e.g., 1/x at 0), or the limit ≠ f(a) (e.g., piecewise jump or removable hole).
Differentiability at x = a: f′(a) exists as a single well-defined value. Requires continuity PLUS smoothness — no corner, no vertical tangent.
Logical chain (one direction only): differentiable ⟹ continuous; not continuous ⟹ not differentiable.
The converse FAILS: continuous ⇏ differentiable; not differentiable ⇏ not continuous. The “corner” case |x| at x = 0 is the standard counter-example.
Modulus functions |f(x)| are continuous wherever f is, but NOT differentiable at the zeros of f where f changes sign — those are the corners.
“Find k so f is continuous”: classic Paper-1 style question. Set lim from below = lim from above = f(a) and solve.
“Find a and b so f is both continuous and differentiable”: two unknowns, two conditions — continuity gives one equation, derivative-matching gives the other.

Continuity at a point — the three-condition test

Don’t take “no holes or jumps” as a wave-of-the-hand — it has a precise three-step test you can actually compute.

Continuity at x = a — all three required (1) f(a) is defined (2) lim_{x → a} f(x) exists (3) lim_{x → a} f(x) = f(a)

Step (2) “the limit exists” itself means both one-sided limits exist AND agree — that’s two sub-conditions hidden in one. So in total you’re really checking four things.

Failure mode 1 — undefined

f(a) doesn’t exist

e.g., f(x) = 1/x is not continuous at x = 0 because f(0) is undefined

Failure mode 2 — value mismatch

lim ≠ f(a)

piecewise with a “jump”; or removable hole filled with the wrong value

Differentiability — continuity plus smoothness

Differentiability is the stronger condition. A function is differentiable at a point if you can draw a unique well-defined tangent line there — which fails whenever the function has a corner (left and right derivatives disagree) or a vertical tangent (derivative is infinite).

The logical chain — one direction only differentiable ⟹ continuous not continuous ⟹ not differentiable
but: continuous ⇏ differentiable not differentiable ⇏ not continuous

Same underlying expression, two functions. Left: x² − 1 is smooth — differentiable for every x. Right: |x² − 1| has corners exactly where x² − 1 changes sign (at x = ±1) — continuous, but not differentiable at those two points.

The modulus rule: |f(x)| is continuous wherever f is, but is NOT differentiable at any zero of f where f changes sign — those zeros become corners after the flip.

🧭 Recipe — test continuity (and differentiability) at x = a

Check f(a) is defined — if undefined, the function fails both continuity and differentiability immediately. STOP.
Compute the two one-sided limits lim_{x → a⁻} and lim_{x → a⁺}. If they disagree, the two-sided limit fails to exist → not continuous → not differentiable. STOP.
Compare the common limit to f(a). If they match, f is continuous. If not (e.g., a removable hole with wrong value), f is not continuous.
For differentiability, also check derivatives: compute the left derivative f′(a⁻) and right derivative f′(a⁺). If equal, differentiable; if not, there’s a corner.
State the conclusion explicitly — name the failure mode (undefined, jump, removable, corner, or vertical tangent) when the test fails.

Worked examples

WE 1

Is the piecewise function continuous at the boundary? (yes-case)

The function f is defined by f(x) = x² + 1 for x ≤ 2, and f(x) = 3x − 1 for x > 2. Use limits to determine whether f is continuous at x = 2.

Step 1 — find f(2). Use the x ≤ 2 piece since 2 satisfies x ≤ 2 f(2) = (2)² + 1 = 5 Step 2 — limit from below (use left piece) lim x→2⁻ (x² + 1) = 4 + 1 = 5 Step 3 — limit from above (use right piece) lim x→2⁺ (3x − 1) = 6 − 1 = 5 Step 4 — compare all three f(2) = 5, lim from below = 5, lim from above = 5 all three equal → f is continuous at x = 2 ✓ f IS continuous at x = 2 when the algebra “happens to work out” so both pieces meet at the same height at the boundary, the function is continuous — even though the two pieces have different formulas.

WE 2

Removable discontinuity — limit exists but disagrees with f(a)

The function g is defined by g(x) = x² + 5x − 6x − 1 for x ≠ 1, and g(1) = 5. Use limits to show that g is not continuous at x = 1.

Step 1 — find g(1). Given directly: g(1) = 5 f(1) = 5 Step 2 — find lim x→1 g(x) using the x ≠ 1 formula at x = 1: (1 + 5 − 6)/(1 − 1) = 0/0 — indeterminate, simplify factor: (x² + 5x − 6) = (x − 1)(x + 6) cancel: g(x) = (x − 1)(x + 6)/(x − 1) = x + 6 for x ≠ 1 lim x→1 (x + 6) = 7 Step 3 — compare lim = 7, g(1) = 5 → 7 ≠ 5 since lim ≠ g(1), g is NOT continuous at x = 1 g is not continuous at x = 1 (removable discontinuity) this is a “removable” discontinuity: if g(1) had been DEFINED as 7 instead of 5, the function would be continuous. The wrong value at x = 1 is the only problem.

WE 3

Find k that makes a piecewise function continuous

The function f is defined by f(x) = x² + k for x < 2, and f(x) = 5x − 3 for x ≥ 2, where k is a real constant. Find the value of k for which f is continuous at x = 2.

Step 1 — find f(2). Use right piece since 2 ∈ [2, ∞) f(2) = 5(2) − 3 = 7 Step 2 — limit from above (using right piece) lim x→2⁺ (5x − 3) = 7 = f(2) ✓ (automatic) Step 3 — limit from below (using left piece, depends on k) lim x→2⁻ (x² + k) = 4 + k Step 4 — set continuity condition need lim from below = f(2) 4 + k = 7 → k = 3 k = 3 makes f continuous at x = 2 paper-1 favourite: a single equation in one unknown drops out of the continuity condition. The right piece is “tied” to f(2) automatically because of how x ≥ 2 is defined.

WE 4

Modulus function — continuous everywhere, but with corners

The function f is defined by f(x) = |x² − 4|. Identify (a) all points (if any) at which f is not continuous, and (b) all points (if any) at which f is not differentiable. Justify briefly.

(a) continuity x² − 4 is continuous everywhere (polynomial) | · | is continuous everywhere → |x² − 4| is continuous everywhere no discontinuities (b) differentiability x² − 4 = 0 at x = ±2; these are the candidates for corners on x < -2: x² − 4 > 0, so f(x) = x² − 4 → f'(x) = 2x on -2 < x < 2: x² − 4 < 0, so f(x) = 4 − x² → f'(x) = -2x on x > 2: x² − 4 > 0, so f(x) = x² − 4 → f'(x) = 2x check derivatives at x = 2 f'(2⁻) = -2(2) = -4; f'(2⁺) = 2(2) = 4 -4 ≠ 4 → not differentiable at x = 2 check derivatives at x = -2 (by symmetry) f'(-2⁻) = 2(-2) = -4; f'(-2⁺) = -2(-2) = 4 -4 ≠ 4 → not differentiable at x = -2 (a) continuous everywhere (b) not differentiable at x = ±2 the sign-change points of f are where the modulus “kinks” — at each kink the left and right derivatives have OPPOSITE signs of equal magnitude, so they cannot agree.

WE 5

Continuous but NOT differentiable at a corner

The function f is defined by f(x) = x² for x ≤ 1, and f(x) = 3x − 2 for x > 1. (a) Show that f is continuous at x = 1. (b) Show that f is NOT differentiable at x = 1.

(a) continuity at x = 1 f(1) = (1)² = 1 (left piece, since 1 ≤ 1) lim x→1⁻ (x²) = 1 lim x→1⁺ (3x − 2) = 3 − 2 = 1 all three equal 1 → continuous ✓ (b) check derivatives at x = 1 on x < 1: f(x) = x² → f'(x) = 2x → f'(1⁻) = 2 on x > 1: f(x) = 3x−2 → f'(x) = 3 → f'(1⁺) = 3 2 ≠ 3 → derivatives don’t agree at x = 1 f is NOT differentiable at x = 1 (corner) f is continuous at x = 1 but NOT differentiable there the textbook example of the converse failure: continuous does NOT imply differentiable. The pieces meet (same height at x=1) but with different slopes — a corner.

WE 6

Find a, b making the function BOTH continuous and differentiable

The function f is defined by f(x) = x² + 2 for x ≤ 1, and f(x) = ax + b for x > 1, where a and b are real constants. Find the values of a and b for which f is differentiable at x = 1.

differentiable ⟹ continuous; need BOTH conditions Continuity condition: lim from below = lim from above = f(1) f(1) = (1)² + 2 = 3 (left piece) lim x→1⁻ (x² + 2) = 3 ✓ (automatic) lim x→1⁺ (ax + b) = a + b ⟹ equation 1: a + b = 3 Differentiability condition: left derivative = right derivative at x = 1 f'(x) = 2x on x < 1 → f'(1⁻) = 2 f'(x) = a on x > 1 → f'(1⁺) = a ⟹ equation 2: a = 2 solve the system a = 2 (from equation 2); sub into equation 1: 2 + b = 3 → b = 1 a = 2 and b = 1 two unknowns, two conditions: continuity pins one relationship between a and b, derivative-matching pins the other. The right-piece line then has slope 2 and y-intercept 1, perfectly tangent to x² + 2 at x = 1.

💡 Top tips

Run the three-condition check in order: f(a) defined? then limit exists? then limit = f(a)? Stop at the first failure.
For piecewise functions, only the boundary is suspect — inside each piece the function is whatever-it-is (polynomial, rational, exponential), inherited from those nice families.
Modulus functions |f(x)| have corners at zeros of f where f changes sign — those are exactly the non-differentiable points.
“Continuous and differentiable” → two equations, normally enough to pin down two parameters (like a and b) in a piecewise definition.
Memorise the four implications (and their two non-implications): differentiable ⟹ continuous; not continuous ⟹ not differentiable; the converses fail.

⚠ Common mistakes

“Continuous because f(a) = lim from below” — you need lim from ABOVE too. Always check both one-sided limits AND the value.
Confusing “f(a) undefined” with “limit doesn’t exist” — they’re different failure modes (and a limit can exist even when f(a) doesn’t).
Assuming “continuous ⟹ differentiable” — false. The corner case (|x| at 0, or WE 5 above) shows continuous without differentiable.
Forgetting to ALSO check continuity when asked about differentiability — differentiability requires continuity, so continuity is a prerequisite check.
For piecewise functions, using the wrong piece for f(a): the equality (≤ or ≥) tells you which piece OWNS the boundary point.

Up next: Further Differentiation. With limits, continuity, and basic derivatives all in place, you’ll learn the advanced HL techniques for differentiating things you can’t easily isolate: implicit differentiation (when y isn’t given explicitly in terms of x), logarithmic differentiation (for ugly products / quotients / variable exponents), derivatives of inverse functions, and the limit definition f′(x) = lim_{h → 0} [f(x+h) − f(x)]/h from first principles.

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