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high school Continuity of Functions

Continuity of Functions

Continuity means you can draw the whole graph in one stroke without lifting your pen off the paper. It is a smooth curve with no sudden breaks, no staircase jumps, and no punched-out holes anywhere. Strictly speaking, a function is continuous at a point only when its limit and its actual value match exactly there. Tap the continuous, jump, and removable buttons and an otherwise smooth graph falls apart in different places.

Intuition of Continuity

What is continuity?
①"A graph you can draw without lifting the pen"
②No breaks, no jumps, no holes — a smooth curve
③Mathematically: three conditions must hold simultaneously

Continuous vs Discontinuous

Three Conditions
continuous at x = a ⟺ ①f(a) exists ②limx→a f(x) exists ③limx→a f(x) = f(a)
If any condition fails ⇒ discontinuous
🔍 Types of Discontinuity
①Jump: left limit ≠ right limit (step function)
②Removable: limit exists but f(a) differs from it (hole)
③Infinite: the limit itself is ±∞ (vertical asymptote)
④The buttons draw continuous / jump / removable only. Infinite discontinuity is text

Properties of Continuous Functions

Arithmetic
If f, g continuous, then f ± g, f · g, f/g (g ≠ 0) are continuous
Sums, products, quotients of continuous functions stay continuous
Composition
f continuous at a, g continuous at f(a) ⇒ g ∘ f continuous at a
Composition of continuous functions is continuous

Intermediate Value Theorem

0.5
IVT
f(a) < k < f(b) ⇒ ∃c ∈ (a, b), f(c) = k
Hypothesis: continuous on [a, b]. Also holds if f(a) > f(b) as long as k is between them. Root-existence tool
💡 Using IVT
①"If f(a) < 0 and f(b) > 0, there is c ∈ (a, b) with f(c) = 0"
②Essential for proving existence of real roots
③f must be continuous on [a, b]

Wrap-up

Definition
limx→a f(x) = f(a)
Limit equals function value ⇒ continuous
🎯 Exam Points
①Three-condition continuity: f(a) exists + limit exists + they're equal
②Polynomials, trig, exp/log are continuous on their domains
③Discontinuity check: compare left and right limits
④IVT: existence of roots — "sign change ⇒ root"
⑤Unknown-coefficient problems: use continuity (left=right=f(a)) to fix constants

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.
1
For continuity at x=1, we need limx→1 f(x) = f(1) = k.
x2 - 1x - 1 = x + 1 (x ≠ 1)
2
Find the limit and set it equal to k.
limx→1 (x + 1) = 2 = k
k = 2
A removable discontinuity becomes continuous when the value equals the limit.
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).
1
f is a polynomial, so it is continuous on [1, 2]; check the signs at the endpoints.
f(1) = -2 < 0, f(2) = 1 > 0
2
The signs differ, so by the IVT there is 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).
If a continuous function changes sign across an interval, a root lies between (IVT).
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.
1
3
2
-1
0
① 1
1
For continuity at x=1, left limit = right limit = f(1).
left = 2(1) + a = 2 + a, right = 12 + 2 = 3
2
Set them equal.
2 + a = 3 ⟹ a = 1
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Derivative and Differentiation
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