Calculus 1A, Lesson 3: Continuity & the Intermediate Value Theorem — interactive worksheet

ELI5: a function is continuous if you can draw it without lifting your pencil — no holes, no sudden jumps, no breaks. Formally, f is continuous at a point when the value it heads toward equals the value it actually has: limit = f(a).

Continuity buys you a beautiful guarantee, the Intermediate Value Theorem (IVT): if a continuous function starts below a target and ends above it, it must hit the target somewhere in between. (You can't climb from the basement to the attic without passing every floor.)

The polynomial x³ − x − 1 is continuous everywhere. At x = 1 it is negative; at x = 2 it is positive. So somewhere between, it is forced to cross zero — a root must exist.

p_{iv}\left(x\right) = x^{3} - x - 1

p_{at_1} = \mathrm{p_{iv}}\left(1\right)

-1

negative: −1

p_{at_2} = \mathrm{p_{iv}}\left(2\right)

5

positive: +5

✓ pass

p_{at_1} < 0 and p_{at_2} > 0

opposite signs ⇒ IVT guarantees a root in (1, 2)

root_{iv} := nsolve("x^3 - x - 1", "x", 1.5)

1.3247

and here it is ≈ 1.3247 — click Steps to replay the hunt

✓ pass

\mathrm{abs}\left(\mathrm{p_{iv}}\left(root_{iv}\right)\right) < 0.000001

plug it back in: genuinely a root

Real-world hook: a thermostat warming a room from 18 °C to 22 °C must pass through every temperature between — including exactly 20 °C. Any continuous signal crossing a threshold is the IVT at work. It's also why numeric solvers like nsolve can find roots at all: bracket a sign change, and a root is guaranteed to be trapped inside.

cbrt_{ans} := nsolve("x^3 - 6", "x", 2)

1.8171

✏️ Your turn: find the cube root of 6 (the x with x³ = 6). Use nsolve("x^3 - 6", "x", 2) — the answer is hidden, so you have to actually find it.

✓ pass

\mathrm{abs}\left(cbrt_{ans}^{3} - 6\right) < 0.0001

green when cubing your answer gives back 6

Where next: Calculus 1B — The Derivative turns "where is it heading?" into "how fast is it changing?"