2.4 Continuity and the Intermediate Value Theorem

2.4 Continuity and the Intermediate Value Theorem

Core CLEP Concept: Continuity is the mathematical formalization of an "unbroken curve." On the CLEP Calculus exam, continuity is tested through three primary question archetypes: verifying the formal 3-part definition at a point, solving for unknown constants in piecewise functions, and applying the Intermediate Value Theorem (IVT) to prove the existence of roots.

---

1. The Formal 3-Part Definition of Continuity

A function $f(x)$ is continuous at a point $x = c$ if and only if all three of the following conditions are simultaneously satisfied:

$$\begin{aligned} \mathbf{1.} & \quad f(c) \text{ is defined } (c \text{ is in the domain of } f). \\ \mathbf{2.} & \quad \lim_{x \to c} f(x) \text{ exists } (\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L \in \mathbb{R}). \\ \mathbf{3.} & \quad \lim_{x \to c} f(x) = f(c) \text{ (The limiting value equals the functional value)}. \end{aligned}$$

   Condition 1: Point Exists        Condition 2: Limits Match        Condition 3: Hole is Plugged
            o (c, f(c))                       o                                *
           / \                               / \                              / \
          /   \                             /   \                            /   \
         /     \                           /     \                          /     \
   -----+-------+----->             -----+-------+----->             -----+-------+----->
                c                                c                                c
   Fails Part 3 (f(c) != L)          Fails Part 1 (f(c) undefined)    Continuous: Part 1, 2, 3 met!

If any one of these three conditions fails, the function is discontinuous at $x = c$.

---

2. Continuity on Intervals (Open and Closed)

• Open Interval $(a, b)$: $f$ is continuous on $(a, b)$ if it is continuous at every individual point $c \in (a, b)$.
• Closed Interval $[a, b]$: $f$ is continuous on $[a, b]$ if:
  1. $f$ is continuous on the open interior $(a, b)$;
  2. $f$ is continuous from the right at $x = a$: $\lim_{x \to a^+} f(x) = f(a)$;
  3. $f$ is continuous from the left at $x = b$: $\lim_{x \to b^-} f(x) = f(b)$.

Continuity of Elementary Functions

The following standard functions are continuous at every number in their algebraic domains:

• Polynomials: Continuous on $(-\infty, \infty)$.
• Rational Functions: Continuous everywhere except where denominator $= 0$.
• Root Functions: $\sqrt[n]{x}$ is continuous on $[0, \infty)$ for even $n$, and $(-\infty, \infty)$ for odd $n$.
• Trigonometric Functions: $\sin(x)$ and $\cos(x)$ are continuous on $(-\infty, \infty)$; $\tan(x)$ is continuous on all $x \neq \frac{\pi}{2} + k\pi$.
• Exponential and Logarithmic Functions: $a^x$ is continuous on $(-\infty, \infty)$; $\ln(x)$ is continuous on $(0, \infty)$.

---

3. Comprehensive Taxonomy of Discontinuities

Discontinuities on the CLEP exam fall into four distinct categories:

Analytical Examples of Each Discontinuity Type

1. Removable Discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$. At $x = 2$, $\lim_{x \to 2} f(x) = 4$, but $f(2)$ is undefined. Redefining $f(2) = 4$ removes the discontinuity.
2. Jump Discontinuity: $f(x) = \frac{|x - 3|}{x - 3}$. $\lim_{x \to 3^-} f(x) = -1$ and $\lim_{x \to 3^+} f(x) = 1$. The jump magnitude is $|1 - (-1)| = 2$.
3. Infinite Discontinuity: $f(x) = \frac{1}{(x - 4)^2}$. As $x \to 4$, $f(x) \to +\infty$.
4. Oscillating Discontinuity: $f(x) = \cos(1/x)$ at $x = 0$.

---

4. Solving for Unknown Parameters in Piecewise Functions

A classic CLEP question requires finding constants that make a piecewise function continuous across transition boundaries.

Master 4-Step Method

1. Set up the left-hand limit $\lim_{x \to c^-} f(x)$ using the left branch.
2. Set up the right-hand limit $\lim_{x \to c^+} f(x)$ using the right branch.
3. Evaluate $f(c)$.
4. Set $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ and solve for the unknown parameter(s).

Step-by-Step Worked Example: Two Unknowns System

Problem: Find constants $a$ and $b$ such that $g(x)$ is continuous for all $x \in \mathbb{R}$:

$$g(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ ax + b & \text{if } 1 \le x \le 3 \\ x^2 - 2 & \text{if } x > 3 \end{cases}$$

Step 1: Ensure continuity at the first boundary $x = 1$.

• $\lim_{x \to 1^-} g(x) = 2(1) + 1 = 3$
• $\lim_{x \to 1^+} g(x) = a(1) + b = a + b$
• $g(1) = a + b$
• Equate: $\mathbf{a + b = 3}$ (Equation 1)

Step 2: Ensure continuity at the second boundary $x = 3$.

• $\lim_{x \to 3^-} g(x) = a(3) + b = 3a + b$
• $\lim_{x \to 3^+} g(x) = 3^2 - 2 = 7$
• $g(3) = 3a + b$
• Equate: $\mathbf{3a + b = 7}$ (Equation 2)

Step 3: Solve the system of linear equations. Subtract Equation 1 from Equation 2:

$$(3a + b) - (a + b) = 7 - 3 \implies 2a = 4 \implies a = 2$$

Substitute $a = 2$ back into Equation 1:

$$2 + b = 3 \implies b = 1$$

Conclusion: $a = 2$ and $b = 1$.

---

5. The Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) is an existence theorem providing guaranteed functional values for continuous functions.

Formal Statement of the IVT

Intermediate Value Theorem: Let $f$ be a function continuous on the closed interval $[a, b]$, and let $N$ be any number strictly between $f(a)$ and $f(b)$, where $f(a) \neq f(b)$. Then there exists at least one number $c \in (a, b)$ such that: $f(c) = N$

   y
   ^
   |                               * (b, f(b))
   |                              /
 f(b) +--------------------------+-
   |                            / |
   N +-------------------------*--+----
   |                          /|  |
 f(a) +------*               / |  |
   |         |\             /  |  |
   |         | \___________/   |  |
   +---------+-----------------+--+-----> x
             a                 c  b
             [<---- c in (a, b) ---->]

Bolzano's Theorem: Guaranteeing Roots (Zeros)

A crucial corollary of the IVT states that if $f(x)$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs ($f(a) \cdot f(b) < 0$), then $0$ lies strictly between $f(a)$ and $f(b)$. Consequently, there exists at least one $c \in (a, b)$ such that:

$$f(c) = 0$$

Step-by-Step Worked Example: Proving Existence of a Root

Problem: Prove that the equation $x^5 - 3x - 1 = 0$ has at least one real solution in the interval $[1, 2]$.

Step 1: Define the function and verify the continuity hypothesis. Let $f(x) = x^5 - 3x - 1$. As a polynomial, $f(x)$ is continuous on $(-\infty, \infty)$ and is therefore continuous on the closed interval $[1, 2]$. (Explicitly stating the continuity hypothesis is mandatory on AP and collegiate calculus exams.)

Step 2: Evaluate the function at the interval endpoints.

• Left endpoint $x = 1$: $$f(1) = 1^5 - 3(1) - 1 = 1 - 3 - 1 = -3 \quad (f(1) < 0)$$
• Right endpoint $x = 2$: $$f(2) = 2^5 - 3(2) - 1 = 32 - 6 - 1 = 25 \quad (f(2) > 0)$$

Step 3: Apply the Intermediate Value Theorem. Since $f(1) = -3 < 0 < 25 = f(2)$, the target value $N = 0$ lies strictly between $f(1)$ and $f(2)$. Because $f$ is continuous on $[1, 2]$, by the IVT there exists at least one number $c \in (1, 2)$ such that $f(c) = 0$.

---

6. Common CLEP Traps with IVT and Continuity

1. Forgetting to Check the Continuity Hypothesis: The IVT fails completely if $f$ has even a single discontinuity in $[a, b]$. For example, consider $f(x) = \frac{1}{x}$ on $[-1, 1]$. Here $f(-1) = -1 < 0$ and $f(1) = 1 > 0$. However, there is no $c \in (-1, 1)$ where $f(c) = 0$ because $f(x)$ is discontinuous at $x = 0$.
2. Assuming the Solution Is Unique: The IVT guarantees that a solution exists (at least one), but it does not guarantee how many solutions exist. A curve may oscillate and cross $y = N$ multiple times.
3. Confusing Open and Closed Intervals: The IVT guarantees $c$ lies strictly in the open interval $(a, b)$, not at the endpoints $a$ or $b$.
4. Assuming the Converse Holds: A discontinuous function may still satisfy the intermediate value property (Darboux functions), but IVT cannot be used to prove continuity.