# Chapter 5: Continuity and Differentiability

> From continuity at a point and on intervals, through IVT and the Extreme Value Theorem, to velocity vs speed, tangent lines, the derivative as a limit of the difference quotient, the second derivative, and when derivatives fail to exist.

**Canonical:** https://calculus.kitzo.work/chapter/continuity  
**Level:** Core  
**Estimated time:** 55 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Continuity at a point
- Continuity on an interval
- Intermediate Value Theorem
- Extreme Value Theorem
- Average vs instantaneous velocity
- Tangent lines
- The derivative as a limit
- Differentiability ⇒ continuity

**Why this matters**

Continuity is what lets calculus trust the world between the points we know, and the derivative is what measures change at a single instant. Every later chapter — rules, applications, integrals — is built on these two ideas.

## 1 · Continuity at a point — the three conditions

f is **continuous at x = a** when all three hold:

  - f(a) is defined,

  - lim x→a f(x) exists,

  - lim x→a f(x) = f(a).

If any fails, f has a **discontinuity** at a: removable (a hole — factor cancels), jump (one-sided limits disagree), or infinite (vertical asymptote). Polynomials are continuous everywhere; rational functions are continuous everywhere except where the denominator vanishes.

## 2 · Continuity on an interval — and the IVT

f is continuous on [a, b] when it is continuous on (a, b) and continuous from the right at a and from the left at b. Continuous-on-an-interval is the hypothesis that unlocks the two big existence theorems:

**Intermediate Value Theorem (IVT):** if f is continuous on [a, b] and N lies between f(a) and f(b), some c in (a, b) has f(c) = N. Sign change between endpoints ⇒ a root inside.

**Worked example**

**Example 1 — Show x³ + x − 1 = 0 has a root in (0, 1).**

Step 1: f(x) = x³ + x − 1 is a polynomial, hence continuous everywhere.

Step 2: f(0) = −1 (negative); f(1) = 1 + 1 − 1 = 1 (positive).

Step 3: 0 lies between −1 and 1, so by the IVT some c in (0, 1) satisfies f(c) = 0.

Notice: the IVT guarantees a root exists without finding it. That is why it is the standard tool for proving existence (and for bisection algorithms on computers).

## 3 · Extreme Value Theorem — max and min exist

**Extreme Value Theorem (EVT):** a continuous function on a closed, bounded interval [a, b] attains an absolute maximum and an absolute minimum there. Both hypotheses matter — on an open interval or with a blow-up, the max/min can escape to the boundary or to infinity. EVT is why optimization problems on [a, b] always have an answer, and why closed intervals are non-negotiable in the fine print.

## 4 · From average velocity to instantaneous velocity

If s(t) is position, **average velocity** over [t, t + h] is Δs/Δt = (s(t + h) − s(t))/h — displacement over time. Shrink h toward 0 and the averages settle onto the **instantaneous velocity**: v(t) = lim h→0 (s(t + h) − s(t))/h. Displacement is change in position (can be negative); total distance traveled adds the absolute values. Speed is |v(t)|, always nonnegative.

## 5 · Tangent lines and the derivative as a limit

The **tangent line** at (a, f(a)) is the line that best matches f's direction there. Its slope is the **derivative**:

`f′(a) = lim h→0 [f(a + h) − f(a)] / h`

Equivalently, `f′(a) = lim x→a [f(x) − f(a)] / (x − a)`. The **derivative function** f′(x) assigns to every x where the limit exists the slope of the tangent there. A linear function f(x) = mx + b has f′(x) = m — constant everywhere, the anchor example for everything that follows.

**Worked example**

**Example 2 — Differentiate f(x) = x² from the definition.**

Step 1: set up the quotient: [(x + h)² − x²] / h.

Step 2: expand: (x² + 2xh + h² − x²)/h = (2xh + h²)/h = 2x + h (valid for h ≠ 0).

Step 3: let h → 0: f′(x) = 2x.

So the tangent to y = x² at x = 3 has slope 6 — the parabola's slope grows without bound as you slide right, and it is exactly 0 at the vertex.

The **second derivative** f″(x) = (f′)′(x) measures the rate of change of the rate of change — acceleration when s is position, or the bending (concavity) of a graph. f″ > 0 means curving upward; f″ < 0 curving downward.

## 6 · When derivatives fail to exist — and the one-way street

Differentiability is *stronger* than continuity. Where f′ fails to exist:

  - **Corners/cusps** — |x| at 0: left slope −1, right slope +1, no single tangent.

  - **Vertical tangents** — x^(1/3) at 0: the tangent is vertical, slope unbounded.

  - **Discontinuities** — jumps and holes never have tangent slopes.

**Differentiability implies continuity**: if f′(a) exists, f must be continuous at a (proof sketch: f(x) − f(a) = [(f(x) − f(a))/(x − a)]·(x − a) → f′(a)·0 = 0). The converse is false — |x| is continuous at 0 but not differentiable there.

**Watch out**

Do not read f′(x) = 0 as "f is flat everywhere" or "f has a maximum" — it only says the tangent is horizontal at that point. Peaks, troughs, plateaus and inflection-level terraces all pass through horizontal tangents.

**Chapter formula sheet**

f continuous at a ⇔ f(a) defined · lim x→a f(x) exists · lim x→a f(x) = f(a)

IVT: f continuous on [a,b], N between f(a) and f(b) ⇒ ∃ c with f(c) = N

EVT: f continuous on [a,b] ⇒ f attains absolute max and min on [a,b]

f′(a) = lim h→0 [f(a + h) − f(a)] / h = lim x→a [f(x) − f(a)] / (x − a)

f linear ⇒ f′ = m · v = s′ · a = v′ = s″ · differentiable ⇒ continuous (not conversely)

## Common mistakes

  - Checking only two of the three continuity conditions — defined, limit exists, and equal are all required.

  - Applying the IVT without verifying continuity, or reading it as "there is exactly one root".

  - Assuming differentiability from continuity — |x| at 0 is the standing counterexample.

  - Computing average velocity and calling it instantaneous; shrink the interval first, or take the limit.

  - Sign errors in the difference quotient: it is f(a + h) − f(a) over h, not f(a) − f(a + h).

  - Thinking f″ > 0 means the function is increasing — it means concave up; f′ is what measures increase.

## Practice exercises

1. f is continuous at x = a when:

- A. f(a) is defined
- B. lim x→a f(x) exists
- C. lim x→a f(x) = f(a)
- D. all three of the above

2. Let f(x) = (x² − 1)/(x − 1) for x ≠ 1, and f(1) = 2. Is f continuous at x = 1? Answer yes or no.

3. The IVT allows you to conclude that x³ + x − 1 = 0 has a root in (0, 1) because:

- A. f(0) and f(1) have opposite signs and f is continuous
- B. f is a polynomial
- C. the function is increasing
- D. the derivative never vanishes

4. A particle has position s(t) = t² + 1 (meters, seconds). What is its average velocity over [1, 3] in m/s?

5. Using f(x) = x² and the limit definition, find f′(3).

6. Which statement is TRUE?

- A. Continuous ⇒ differentiable
- B. Differentiable ⇒ continuous
- C. Both are equivalent
- D. Neither implies the other

7. Compute the derivative of f(x) = 5x + 7 (linear function). Enter f′(x).

8. At a corner point such as f(x) = |x| at x = 0, which condition for differentiability fails?

- A. f is not continuous there
- B. The left- and right-hand difference quotients disagree
- C. f(0) is undefined
- D. The limit of f(x) as x → 0 does not exist

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