Lesson 4: Continuity & the IVT

(a) Opening Question

Consider the function

        ⎧ x² - 4
        ⎪ ───────,   x ≠ 2
f(x) =  ⎨ x - 2
        ⎪
        ⎩ 5,         x = 2

Two quick questions before we begin.

1. What is lim_{x→2} f(x)? (Hint: factor the numerator. You did exactly this kind of cancellation in Lesson 3.)
2. Does that limit equal the value f(2) that the rule actually assigns?

Work it out before reading on. Factoring, (x² - 4)/(x - 2) = (x-2)(x+2)/(x-2) = x + 2 for every x ≠ 2, so lim_{x→2} f(x) = 2 + 2 = 4. But the function was defined to equal 5 at x = 2. So the limit (4) and the function value (5) disagree.

The graph is the line y = x + 2 with the single point at x = 2 lifted up to height 5 — a hole at (2, 4) and a stray dot at (2, 5). That tiny mismatch is exactly what "continuity" measures, and repairing it is the heart of this lesson.

(b) Core Concepts

Continuity at a point: the three conditions

Informally, a function is continuous if you can draw its graph through that point without lifting your pencil. To make that precise, the AP exam requires you to check three separate conditions.

A function f is continuous at x = c if and only if all three hold:

(1)  f(c) is defined            (c is in the domain)
(2)  lim_{x→c} f(x) exists      (left-hand limit = right-hand limit, both finite)
(3)  lim_{x→c} f(x) = f(c)      (the limit equals the function value)

These are ordered for a reason. Condition (2) requires the one-sided limits to agree (recall Lesson 2). Condition (3) then requires that common limiting value to actually match the height of the dot. If any one of the three fails, f is discontinuous at c.

In the Opening Question, conditions (1) and (2) both held — f(2) = 5 is defined, and lim_{x→2} f(x) = 4 exists — but condition (3) failed because 4 ≠ 5. That single failure is enough.

Continuity on an interval

f is continuous on an open interval (a, b) if it is continuous at every point inside it. For a closed interval [a, b], we only have one side available at the endpoints, so we require continuity at every interior point plus one-sided continuity at the ends:

lim_{x→a⁺} f(x) = f(a)      and      lim_{x→b⁻} f(x) = f(b)

This closed-interval idea matters enormously, because the Intermediate Value Theorem (and later the Extreme Value Theorem and the Mean Value Theorem) all demand continuity on a closed interval [a, b].

The three types of discontinuity

When f fails to be continuous at c, the kind of failure has a name.

• Removable discontinuity (a hole). lim_{x→c} f(x) exists (a finite number), but either f(c) is undefined or f(c) does not equal that limit. We could "remove" it by redefining f(c) to be the limit. (Both the Opening Question and any cancelled factor like (x²-4)/(x-2) produce removable discontinuities.)
• Jump discontinuity. The left-hand and right-hand limits both exist but are not equal, so condition (2) fails. The graph "jumps" from one height to another. No single value of f(c) can fix this.
• Infinite discontinuity. At least one one-sided limit is +∞ or −∞, typically at a vertical asymptote such as f(x) = 1/(x - c).

[GRAPH: three side-by-side panels illustrating discontinuity types, each on [-1, 5] × [-4, 4]

Panel 1 — REMOVABLE (hole): the line y = x - 1 drawn through the window, with an open circle (hole) at the point (2, 1). A separate filled dot sits at (2, 3), above the hole. Label: "limit exists (= 1) but f(2) ≠ limit".

Panel 2 — JUMP: a step function. For x < 2 draw the horizontal segment y = 1 ending in an open circle at (2, 1). For x ≥ 2 draw the horizontal segment y = 3 starting with a filled dot at (2, 3). Label: "left limit = 1, right limit = 3; limits disagree".

Panel 3 — INFINITE: the curve y = 1/(x - 2) with a dashed vertical asymptote at x = 2. Branch on the left falls to -∞ as x→2⁻; branch on the right rises to +∞ as x→2⁺. Label: "one-sided limit is infinite".]

Which functions are automatically continuous?

You do not re-derive continuity from scratch every time. These common function families are continuous at every point of their domains:

• Polynomials — continuous everywhere.
• Rational functions p(x)/q(x) — continuous everywhere except where q(x) = 0.
• Root functions like √x — continuous on their domain (x ≥ 0).
• Exponential eˣ, aˣ — continuous everywhere.
• Logarithmic ln x — continuous for x > 0.
• Trigonometric sin x, cos x — continuous everywhere; tan x, sec x etc. continuous except where undefined.

Sums, differences, products, quotients (nonzero denominator), and compositions of continuous functions are continuous. So f(x) = e^{x} · cos x + √(x² + 1) is continuous for all real x without any further work — a fact you can state in a justification.

Using continuity to find an unknown parameter

A favorite AP setup gives a piecewise function with an unknown constant and asks you to choose the constant that makes the function continuous. The strategy is always the same: at the boundary point, force condition (3) by setting the two pieces equal.

Mini-example. Let g(x) = kx + 1 for x ≤ 3 and g(x) = x² - 5 for x > 3. Find k making g continuous at x = 3.

Each piece is a polynomial, so the only place continuity can fail is the boundary x = 3. We need the left and right limits to match the value:

- Left piece at 3: k(3) + 1 = 3k + 1

- Right limit as x→3⁺: (3)² - 5 = 4

Set them equal: 3k + 1 = 4 ⟹ 3k = 3 ⟹ k = 1. With k = 1, both pieces give 4 at x = 3, all three conditions hold, and g is continuous.

The Intermediate Value Theorem

Continuity is what lets us guarantee a function hits a target value somewhere, even when we cannot solve for the exact location.

Intermediate Value Theorem (IVT).

Hypotheses: If f is continuous on the closed interval [a, b], and N is any number between f(a) and f(b) (with f(a) ≠ f(b)),

Conclusion: then there exists at least one number c in the open interval (a, b) such that f(c) = N.

Read the hypotheses carefully — the AP exam scores them:

1. Continuous on a closed interval. Both endpoints included; no gaps, jumps, or asymptotes anywhere in [a, b]. If f is discontinuous, the theorem says nothing.
2. N is between the endpoint values. It must lie strictly between f(a) and f(b). The IVT does not say f hits values outside that range.

The IVT is an existence theorem: it guarantees that some c exists, but it does not tell you what c is or how many there are.

Worked: showing a root exists (model IVT justification)

Show that f(x) = x³ + x - 1 has a root on [0, 1].

A "root" means f(c) = 0, so we take N = 0 and check the endpoints:

f(0) = 0³ + 0 - 1 = -1        (negative)
f(1) = 1³ + 1 - 1 =  1        (positive)

Since −1 < 0 < 1, the target N = 0 lies between f(0) and f(1). Now the full justification in AP language:

f(x) = x³ + x − 1 is a polynomial, so it is continuous on the closed interval [0, 1]. Since f(0) = −1 < 0 < 1 = f(1), the value 0 lies between f(0) and f(1). Therefore, by the Intermediate Value Theorem, there exists at least one value c in (0, 1) such that f(c) = 0.

Notice every hypothesis is named explicitly: continuity, closed interval, N between the endpoint values. That sentence earns full credit; "by IVT there's a root" does not.

(c) Worked Examples

Example 1 — Classify the discontinuities (foundation) · [NO CALC]

Problem. For f(x) = (x² - x - 6)/(x - 3), locate and classify every discontinuity.

Strategy. A rational function is discontinuous only where the denominator is 0. Factor and see whether the factor cancels (removable) or not (infinite).

Solution. The denominator is 0 at x = 3. Factor the numerator:

x² - x - 6 = (x - 3)(x + 2)

So for x ≠ 3, f(x) = (x-3)(x+2)/(x-3) = x + 2. The (x − 3) cancels, so lim_{x→3} f(x) = 3 + 2 = 5, a finite limit, but f(3) is undefined.

Justification. The limit lim_{x→3} f(x) = 5 exists but f(3) is undefined, so condition (1) fails while the limit exists — this is a removable discontinuity (hole) at (3, 5). There are no other discontinuities.

Example 2 — Parameter making a piecewise function continuous (procedural) · [NO CALC]

Problem. Find the value of a for which

        ⎧ x² + a,      x < 1
h(x) =  ⎨
        ⎩ 4x - 1,      x ≥ 1

is continuous at x = 1.

Strategy. Each piece is a polynomial, continuous on its own. Continuity can only break at the boundary x = 1, so enforce condition (3): the left-hand limit must equal h(1).

Solution.

lim_{x→1⁻} h(x) = (1)² + a = 1 + a
h(1) = 4(1) - 1 = 3      (also the right-hand limit)

Set them equal: 1 + a = 3 ⟹ a = 2.

Justification. With a = 2, lim_{x→1⁻} h(x) = 3 = lim_{x→1⁺} h(x), so the two-sided limit exists and equals h(1) = 3. All three conditions hold, so h is continuous at x = 1.

Example 3 — IVT to guarantee a value (AP level) · [NO CALC]

Problem. Let f be continuous on [1, 5] with f(1) = −3 and f(5) = 7. Explain why there must be a value c in (1, 5) with f(c) = 2.

Strategy. This is a direct IVT application. Confirm the hypotheses, identify N = 2, and write the precise justification.

Solution / Justification.

f is continuous on the closed interval [1, 5] (given). Since f(1) = −3 and f(5) = 7, and −3 < 2 < 7, the value N = 2 lies between f(1) and f(5). Therefore, by the Intermediate Value Theorem, there exists at least one value c in (1, 5) such that f(c) = 2.

Note what we did not claim: we did not say where c is, nor that it is unique. The IVT only guarantees existence.

Example 4 — IVT with a calculator solve (AP level) · [CALC]

Problem. Show that f(x) = cos x − x has a solution to f(x) = 0 on [0, 1], then approximate it.

Strategy. Use the IVT for existence (by hand), then use the calculator's numerical solver for the location.

Solution. First the existence justification:

f(0) = cos 0 - 0 = 1        (positive)
f(1) = cos 1 - 1 ≈ 0.5403 - 1 = -0.4597     (negative)

f(x) = cos x − x is continuous on [0, 1] because it is the difference of cos x (continuous everywhere) and the polynomial x. Since f(0) = 1 > 0 > −0.4597 ≈ f(1), the value 0 lies between f(0) and f(1). By the Intermediate Value Theorem, there exists c in (0, 1) with f(c) = 0.

Now locate it:

TI-84: y1 = cos(X) - X.  Graph on [0,1]×[-1,1] to see the single crossing.
       Then 2nd → TRACE (CALC) → 2:zero, bracket the crossing.
       Result: c ≈ 0.739

So c ≈ 0.739 (this is the famous Dottie number, where cos x = x). The IVT guaranteed the crossing existed; the calculator only pinned it down.

(d) Common Mistakes

Checking only the limit (or only the value), not all three conditions. Students see that lim_{x→c} f(x) exists and declare f continuous. Why it's wrong: the limit can exist while f(c) is undefined or unequal (the Opening Question: limit 4, value 5). Fix: every time, verify (1) f(c) defined, (2) limit exists, (3) they are equal.

Applying the IVT without confirming continuity. Students compute f(a) and f(b), see N between them, and conclude a c exists. Why it's wrong: the IVT is false without continuity — a jump function can skip right over N. Fix: state the continuity hypothesis first ("f is continuous on [a, b] because…") before citing the theorem.

Confusing the IVT with the MVT or the EVT. Why it's wrong: the IVT guarantees an output value f(c) = N; the Mean Value Theorem guarantees a slope f′(c) = [f(b)−f(a)]/(b−a) and needs differentiability; the Extreme Value Theorem guarantees a max and a min exist on [a, b]. Fix: IVT → a value is attained. MVT → an instantaneous rate matches the average rate. EVT → extremes exist.

Using the closed interval in the conclusion. Students write "there exists c in [a, b]." Why it's wrong: the hypotheses use the closed interval [a, b], but the conclusion places c in the open interval (a, b). Fix: hypotheses closed, conclusion open.

Forgetting N must be strictly between the endpoint values. Students try to use the IVT to reach a value outside [f(a), f(b)]. Why it's wrong: the theorem only delivers values between f(a) and f(b). Fix: always confirm f(a) < N < f(b) (or f(b) < N < f(a)).

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## (e) Practice Problems

For what value of k is the function below continuous at x = 2?

f(x) = 3x + k   for x ≤ 2,    f(x) = x² + 2   for x > 2

(A) k = 0 (B) k = -1 (C) k = 2 (D) k = 6

(Justification.) Let f(x) = x⁵ + 2x − 5. Show that f has a real root on [1, 2]. Write a complete IVT justification, naming every hypothesis.

Find all discontinuities of f(x) = (x² - 9)/(x² - 4x + 3) and classify each.

Determine constants a and b so that the function is continuous everywhere:

        ⎧ -2,            x ≤ -1
f(x) =  ⎨ ax + b,        -1 < x < 3
        ⎩ 10,            x ≥ 3

(Justification.) A continuous function models the temperature T(t) (in °F) of a metal rod, with T(0) = 68 and T(10) = 200. Use the IVT to justify that at some time the rod's temperature was exactly 150 °F. Write the full justification.

Is f(x) = |x − 2| continuous at x = 2? Justify using the three conditions. (It is differentiable there?)

Explain in one sentence why the IVT cannot be applied to f(x) = 1/x on [−1, 1] to conclude f(c) = 0 for some c, even though f(−1) = −1 < 0 < 1 = f(1).

For f(x) = ln x − 1, use the IVT to justify a solution to f(x) = 0 on [2, 3], then approximate it with a calculator.

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TI-84: y1 = ln(X) - 1.  Graph on [2,3]×[-0.5,0.5] → 2nd TRACE → 2:zero → c ≈ 2.718