Why Continuity Matters
Continuous functions on closed intervals behave very nicely. They don't have gaps, they don't blow up to infinity, and they hit every value in between their endpoints. These properties are captured by two major theorems: the Intermediate Value Theorem (IVT) and the Extreme Value Theorem (EVT).
Intermediate Value Theorem (IVT)
If \( f \) is continuous on \([a, b]\) and \( y \) is a value strictly between \( f(a) \) and \( f(b) \), then there exists at least one \( c \in (a, b) \) such that \( f(c) = y \).
Drag the horizontal line (target value \(y\)) to see where it intersects the function.
Intuition: You cannot get from point A to point B without crossing every horizontal line in between, unless you "teleport" (which would mean the function is discontinuous).
Extreme Value Theorem (EVT)
If \( f \) is continuous on a closed, bounded interval \([a, b]\), then \( f \) is bounded and attains its maximum and minimum values.
Explore why the conditions "closed" and "bounded" are necessary.
Scenario
Practice Problems
True/False
1. If \( f \) is continuous on an open interval \( (a, b) \), then \( f \) is bounded on \( (a, b) \).
False. Consider \( f(x) = 1/x \) on \( (0, 1) \). It is continuous but unbounded near 0.
2. If \( f \) is continuous on \( [a, b] \) and \( f(a) \lt 0 \lt f(b) \), then there exists \( c \in (a, b) \) such that \( f(c) = 0 \).
True. This is a direct application of the Intermediate Value Theorem with \( k=0 \).
3. If \( f \) is continuous on \( [a, b] \), then \( f \) attains a maximum and a minimum on \( [a, b] \).
True. This is the statement of the Extreme Value Theorem.
4. If \( f \) is continuous on \( [a, b] \), then the image set \( f([a, b]) \) is a closed interval.
True. By EVT, let \( m = \min f \) and \( M = \max f \). By IVT, \( f \) takes all values between \( m \) and \( M \). Thus \( f([a, b]) = [m, M] \).
Fill in the Blank
1. The Intermediate Value Theorem states that if \( f \) is continuous on \( [a, b] \) and \( k \) is any number strictly between \( f(a) \) and \( f(b) \), then there exists \( c \in (a, b) \) such that \( f(c) = \) .
Answer: \( k \)
2. The Extreme Value Theorem requires the domain interval to be both and .
Answer: closed, bounded (or compact)
3. To prove that a continuous function \( f: [0, 1] \to [0, 1] \) has a fixed point (i.e., \( f(c) = c \)), we apply the IVT to the function \( g(x) = \) .
Answer: \( f(x) - x \). Note that \( g(0) = f(0) \ge 0 \) and \( g(1) = f(1) - 1 \le 0 \).
Full Problems
1. Show that the equation \( x^3 - 4x + 1 = 0 \) has a solution in the interval \( [0, 1] \).
Let \( f(x) = x^3 - 4x + 1 \). \( f \) is a polynomial, so it is continuous everywhere.
Evaluate at endpoints:
- \( f(0) = 0^3 - 4(0) + 1 = 1 \gt 0 \)
- \( f(1) = 1^3 - 4(1) + 1 = -2 \lt 0 \)
Since \( f(0) \gt 0 \) and \( f(1) \lt 0 \), by the Intermediate Value Theorem, there exists \( c \in (0, 1) \) such that \( f(c) = 0 \).
2. Give an example of a continuous function on \( (0, 1) \) that is not bounded. Explain why this does not contradict the Extreme Value Theorem.
Example: \( f(x) = \frac{1}{x} \). As \( x \to 0^+ \), \( f(x) \to \infty \), so it is not bounded above.
This does not contradict the EVT because the interval \( (0, 1) \) is not closed (it is not compact). The EVT requires a closed and bounded interval \( [a, b] \).
3. Let \( f \) be a continuous function on \( [a, b] \). Prove that if \( f(x) \in [a, b] \) for all \( x \in [a, b] \), then \( f \) has a fixed point.
Let \( g(x) = f(x) - x \). We want to find \( c \) such that \( g(c) = 0 \).
Since \( f(x) \in [a, b] \), we know \( a \le f(x) \le b \) for all \( x \).
- \( g(a) = f(a) - a \ge a - a = 0 \)
- \( g(b) = f(b) - b \le b - b = 0 \)
If \( g(a)=0 \) or \( g(b)=0 \), we are done. Otherwise \( g(a) \gt 0 \) and \( g(b) \lt 0 \). By IVT, there exists \( c \in (a, b) \) such that \( g(c) = 0 \), which implies \( f(c) = c \).
4. Is the function \( f(x) = \frac{x^2}{1+x^2} \) bounded on \( \mathbb{R} \)? Does it attain a maximum?
Yes, it is bounded. \( 0 \le x^2 \lt 1+x^2 \), so \( 0 \le f(x) \lt 1 \) for all \( x \).
It does NOT attain a maximum. The value 1 is the supremum (limit as \( x \to \pm\infty \)), but \( f(x) \) is never equal to 1 for any finite \( x \). This shows that EVT fails if the interval is unbounded (\( \mathbb{R} \)).