Lesson 4: Continuity and the Intermediate Value Theorem
Estimated time: 30-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define continuity at a point using the three-condition test
- Identify and classify types of discontinuities (removable, jump, infinite)
- Determine intervals on which a function is continuous
- State and apply the Intermediate Value Theorem
- Use the IVT to show that an equation has a solution in a given interval
Continuity at a Point
Informally, a function is continuous if you can draw its graph without lifting your pencil. Formally, continuity at a point requires three conditions to hold simultaneously.
Continuity at a Point
A function f is continuous at x = a if all three conditions are satisfied:
- f(a) is defined (a is in the domain of f)
- limx→a f(x) exists
- limx→a f(x) = f(a)
If any condition fails, f is discontinuous at x = a.
Example 1: Checking Continuity
Is f(x) = (x² − 1)/(x − 1) continuous at x = 1?
Step 1: f(1) = 0/0 — undefined. Condition 1 fails!
Conclusion: f is discontinuous at x = 1 because f(1) is not defined.
(Note: the limit limx→1 f(x) = 2 exists, but since f(1) is undefined, continuity fails.)
Example 2: Continuous Polynomial
Show that f(x) = 3x² − x + 5 is continuous at x = 2.
Step 1: f(2) = 3(4) − 2 + 5 = 15. Defined. ✓
Step 2: limx→2 (3x² − x + 5) = 15. Exists. ✓
Step 3: lim = f(2) = 15. ✓
Conclusion: f is continuous at x = 2. (In fact, all polynomials are continuous everywhere.)
Types of Discontinuities
Removable Discontinuity (Hole)
The limit exists at x = a, but either f(a) is undefined or f(a) ≠ limx→a f(x). The graph has a "hole" that could be filled by redefining f(a).
Example: f(x) = (x² − 4)/(x − 2) at x = 2.
Jump Discontinuity
Both one-sided limits exist but are not equal: limx→a− f(x) ≠ limx→a+ f(x). The graph "jumps" at x = a.
Example: The greatest integer function ⌊x⌋ at every integer.
Infinite Discontinuity
At least one one-sided limit is ±∞. The graph has a vertical asymptote at x = a.
Example: f(x) = 1/x at x = 0.
Continuity on an Interval
Continuity on an Interval
f is continuous on (a, b) if it is continuous at every point in (a, b).
f is continuous on [a, b] if it is continuous on (a, b), continuous from the right at a, and continuous from the left at b.
Example 3: Finding Intervals of Continuity
Where is f(x) = 1/(x − 3) continuous?
Solution: f is a rational function. It is continuous wherever the denominator is nonzero, i.e., x ≠ 3.
f is continuous on (−∞, 3) and (3, ∞).
Functions That Are Continuous on Their Domains
The following are continuous at every point in their domains:
- Polynomials (continuous everywhere)
- Rational functions
- Root functions
- Trigonometric functions
- Exponential and logarithmic functions
Building Continuous Functions
If f and g are continuous at a, then so are:
- f + g, f − g, f · g, and c · f (for any constant c)
- f/g (provided g(a) ≠ 0)
- f(g(x)) (the composition, provided g is continuous at a and f is continuous at g(a))
Example 4: Composition Continuity
Is h(x) = sin(x² + 1) continuous everywhere?
Solution: g(x) = x² + 1 is a polynomial (continuous everywhere). f(u) = sin u is continuous everywhere. The composition f(g(x)) = sin(x² + 1) is continuous everywhere. Yes.
The Intermediate Value Theorem (IVT)
Intermediate Value Theorem (IVT)
If f is continuous on the closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N.
In plain language: a continuous function cannot "skip" a value. If f goes from 2 to 5 on an interval, it must hit every value between 2 and 5 at least once.
Example 5: Using IVT to Locate a Root
Show that the equation x³ − x − 1 = 0 has a solution between x = 1 and x = 2.
Step 1: Let f(x) = x³ − x − 1. This is a polynomial, so it is continuous everywhere.
Step 2: Evaluate at the endpoints:
- f(1) = 1 − 1 − 1 = −1 < 0
- f(2) = 8 − 2 − 1 = 5 > 0
Step 3: Since f(1) < 0 < f(2) and f is continuous on [1, 2], the IVT guarantees that there exists some c in (1, 2) with f(c) = 0.
Conclusion: The equation has at least one solution in (1, 2).
Example 6: Narrowing Down a Root
Using the function from Example 5, narrow the root to a smaller interval.
Step 1: Try the midpoint x = 1.5: f(1.5) = 3.375 − 1.5 − 1 = 0.875 > 0.
Step 2: Since f(1) = −1 < 0 and f(1.5) = 0.875 > 0, the root is in (1, 1.5).
Step 3: Try x = 1.25: f(1.25) = 1.953 − 1.25 − 1 = −0.297 < 0.
Step 4: Since f(1.25) < 0 and f(1.5) > 0, the root is in (1.25, 1.5).
This process (the bisection method) can continue to approximate the root to any desired accuracy.
Piecewise Functions and Continuity
Example 7: Making a Piecewise Function Continuous
Find the value of k that makes f continuous at x = 2, where:
f(x) = { kx + 1 if x ≤ 2, and x² − 1 if x > 2 }.
Step 1: For continuity at x = 2, we need limx→2− f(x) = limx→2+ f(x) = f(2).
Step 2: Left-hand limit: limx→2− (kx + 1) = 2k + 1.
Step 3: Right-hand limit: limx→2+ (x² − 1) = 4 − 1 = 3.
Step 4: f(2) = k(2) + 1 = 2k + 1 (using the first piece since x ≤ 2).
Step 5: Set them equal: 2k + 1 = 3, so 2k = 2, so k = 1.
Key Takeaways
- A function is continuous at x = a when three conditions hold: f(a) exists, the limit exists, and they are equal.
- Discontinuities come in three types: removable (holes), jump, and infinite (vertical asymptotes).
- Polynomials, trig functions, exponentials, and logarithms are continuous on their entire domains.
- Sums, products, quotients, and compositions of continuous functions are continuous (where defined).
- The IVT says a continuous function on [a, b] takes on every value between f(a) and f(b).
- The IVT is used to prove that equations have solutions and to approximate roots via bisection.
Check Your Understanding
1. State the three conditions for continuity at x = a.
2. Classify the discontinuity of f(x) = |x|/x at x = 0.
3. Use the IVT to show that ex = 3 − x has a solution on [0, 1].
4. Find k so that g(x) = { x² + k if x < 1, and 3x if x ≥ 1 } is continuous at x = 1.