Grade 11-12 (age 16-18)

high school Continuity of Functions

A function is continuous when its graph can be drawn in one go without lifting the pen from the paper. There must be no breaks, no step-like jumps and no holes along the way. Precisely, f is continuous at x = a when the value f(a) and the limit both exist and are equal. The three buttons in the second picture compare a continuous function, a jump discontinuity and a removable discontinuity; in the fourth, changing the height k finds the point c that the intermediate value theorem promises.

Intuition of Continuity

✏ What continuity means

①A graph you can draw without lifting the pen

②No breaks, no jumps, no holes

③The three conditions below turn that feeling into a definition

Continuous and discontinuous side by side

The three conditions for continuity

continuous at x = a ⟺ ①f(a) exists ②lim_x→a f(x) exists ③lim_x→a f(x) = f(a)

If any one of the three fails, the function is discontinuous

🔍 Kinds of discontinuity

①Jump discontinuity: the left- and right-hand limits differ, so the graph jumps

②Removable discontinuity: the limit exists but f(a) is missing or different, leaving a hole

③Infinite discontinuity: the limit runs off to plus or minus infinity, with a vertical asymptote (not shown in the picture)

④A removable discontinuity becomes continuous once f(a) is redefined as the limit

Properties of Continuous Functions

Arithmetic

If f, g continuous, then f ± g, f · g, f/g (g ≠ 0) are continuous

Sums, differences and products of continuous functions are continuous, and quotients are continuous wherever the denominator is not 0

Composition

f continuous at a, g continuous at f(a) ⇒ g ∘ f continuous at a

A composition of continuous functions is continuous

Intermediate Value Theorem

Height k0.5

IVT

f(a) < k < f(b) ⇒ ∃c ∈ (a, b), f(c) = k

It holds when f is continuous on [a, b]; even if f(a) > f(b), it works for any k between the two values

💡 Using the intermediate value theorem

①If f(a) < 0 < f(b), there is at least one c in (a, b) with f(c) = 0

②That is how you show an equation has a real root in a given interval

③If f is not continuous on [a, b], the theorem cannot be used

④The theorem only says that a c exists, not how many there are; in the picture, any k between −1 and 3 gives three dots

Summary

Definition

lim_x→a f(x) = f(a)

Continuous means the limit equals the function value

🎯 Exam Points

①Three conditions: f(a) exists, the limit exists, and they are equal

②Polynomials are continuous everywhere; rational functions wherever the denominator is not 0

③To test for discontinuity, compare the left- and right-hand limits first

④Intermediate value theorem: if a continuous function changes sign, there is a root in between

⑤For unknown constants, set left limit = right limit = f(a)

Worked Examples & Past Exam

Example 1

For f(x) = (x² − 1)/(x − 1) (x ≠ 1) and f(1) = k, find the constant k that makes f continuous at x = 1.

For continuity at x = 1, the limit must equal f(1) = k.

x² - 1x - 1 = x + 1 (x ≠ 1)

For x ≠ 1 the fraction simplifies to x + 1, so the limit is 2.

lim_x→1 (x + 1) = 2 = k

▸ k = 2

The original function is the line y = x + 1 with a hole at x = 1. Filling the hole with k = 2 makes it continuous.

Example 2

For f(x) = x² − 3, explain that the equation f(x) = 0 has at least one real root in the interval (1, 2).

f is a polynomial, so it is continuous on [1, 2]; check the signs of the values at the ends.

f(1) = -2 < 0, f(2) = 1 > 0

The signs differ, so by the intermediate value theorem there is a c in (1, 2) with f(c) = 0.

f(1) < 0 < f(2) ⟹ f(c) = 0, c ∈ (1, 2)

▸ A real root exists in (1, 2).

The actual root is √3 ≈ 1.73. The intermediate value theorem alone does not give the value of the root, but it does tell you which interval it lies in.

2022 KICE mock exam Math type, adapted

For f(x) = { 2x + a (x ≤ 1), x² + 2 (x > 1) }, find the constant a if f is continuous at x = 1.

①1

②3

③2

④-1

⑤0

▸ ① 1

For continuity at x = 1, the left-hand limit, the right-hand limit and f(1) must all be equal.

left = 2(1) + a = 2 + a, right = 1² + 2 = 3

Set the left-hand limit 2 + a equal to the right-hand limit 3.

2 + a = 3 ⟹ a = 1