Continuity and Differentiability of Functions and Area Functions

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Continuity and Differentiability of Functions and Area Functions

TL;DR

A function is continuous if its graph has no breaks or jumps; it's differentiable if it has a well-defined slope everywhere. Area functions (antiderivatives) are always continuous, but they're only differentiable where the original function is continuous—this is the Fundamental Theorem of Calculus in action.

1. The Mental Model

Think of continuity as "no broken pencil line"—you can draw the graph without lifting your pen. Differentiability is stricter: the graph must be smooth, with no sharp corners or vertical tangents. An area function F(x) tracks cumulative area under a curve; it's always continuous even if the curve below it jumps around, but F only has a slope (F'(x)) where the original curve behaves nicely. The Fundamental Theorem ties these together: the slope of the area function is the height of the original function.

2. The Core Material

Continuity: The Formal Definition

A function f is continuous at x = a if three things happen:
1. f(a) exists (the function has a value there).
2. $\lim_{x \to a} f(x)$ exists (the left and right limits agree).
3. $\lim_{x \to a} f(x) = f(a)$ (the limit equals the actual value).

If all three hold at every point in an interval, f is continuous on that interval. Graphically, you can draw it without lifting your pen.

Why it matters: Continuous functions have nice properties. The Intermediate Value Theorem says if f is continuous on [a, b] and f(a) ≠ f(b), then f takes every value between them. This is why continuous functions on closed intervals are always bounded and always achieve their maximum and minimum.

Differentiability: Slopes Everywhere

A function f is differentiable at x = a if the derivative
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
exists and is finite. This limit is the slope of the tangent line at that point.

Key truth: differentiability is stronger than continuity. If f is differentiable at a, then f is continuous at a. But the reverse is false—a continuous function might have a sharp corner or cusp where it's not differentiable.

Example: f(x) = |x| is continuous everywhere. At x = 0, the left derivative is −1 and the right derivative is +1. They don't match, so f is not differentiable at x = 0 (there's a sharp corner).

Area Functions and the Fundamental Theorem

An area function is typically defined as
$$F(x) = \int_a^x f(t) \, dt$$
where f is some given function and a is a fixed starting point. This accumulates signed area under f from a to x.

The crucial insight: F(x) is always continuous, even if f has jumps. Why? Because the accumulated area can't jump—it changes smoothly as x moves.

The Fundamental Theorem of Calculus (Part 1) says: if f is continuous at x = c, then F is differentiable at c and
$$F'(c) = f(c)$$

Translation: the slope of the area function equals the height of the original function. This is profound—it says integration and differentiation are inverse operations.

Important caveat: If f has a jump discontinuity at some point, then F is still continuous there, but F' doesn't exist at that point. The area function has a "kink" where the original function jumps.

graph TD
    A["Function f is continuous<br/>at x = c"] -->|FTC Part 1| B["Area function F<br/>is differentiable at c"]
    B --> C["F'c = fc<br/>slope of area = height of curve"]
    D["Function f has<br/>jump at x = c"] -->|Still true| E["Area function F<br/>is continuous at c"]
    E -->|But NOT| F["F is differentiable at c<br/>there's a kink"]

3. Worked Example

Problem: Let $f(x) = \begin{cases} 2 & \text{if } x < 1 \\ 5 & \text{if } x \geq 1 \end{cases}$ (a jump discontinuity at x = 1).

Define $F(x) = \int_0^x f(t) \, dt$. Find F(x) on [0, 3], and identify where F is differentiable.

Solution:

For 0 ≤ x < 1: $F(x) = \int_0^x 2 \, dt = 2x$

For x ≥ 1: $F(x) = \int_0^1 2 \, dt + \int_1^x 5 \, dt = 2(1) + 5(x - 1) = 5x - 3$

So $$F(x) = \begin{cases} 2x & \text{if } x < 1 \\ 5x - 3 & \text{if } x \geq 1 \end{cases}$$

Check continuity at x = 1:
- Left limit: $\lim_{x \to 1^-} 2x = 2$
- Right limit: $\lim_{x \to 1^+} (5x - 3) = 2$
- F(1) = 5(1) − 3 = 2

All three agree, so F is continuous at x = 1. ✓

Check differentiability at x = 1:
- Left derivative: F'(1−) = 2
- Right derivative: F'(1+) = 5

They don't match, so F is not differentiable at x = 1. The area function has a corner.

Elsewhere: For x < 1 and x > 1, both sides of F are linear, so F' exists and equals 2 or 5 respectively. Notice F'(x) = f(x) wherever f is continuous (on (−∞, 1) and (1, ∞)).

4. Key Takeaways

4.1 Most Important Concepts

1. Continuity = no breaks. The function value, the limit, and they match at every point.

2. Differentiability = smooth, no corners. Requires continuity plus the slope from the left equals the slope from the right.

3. Differentiability implies continuity. Not vice versa: |x| is continuous but not differentiable at 0.

4. Area functions are always continuous. Even if f is wild, $\int_a^x f(t) \, dt$ has no jumps.

5. The Fundamental Theorem connects them. If f is continuous, the area function's slope is f.

6. Jumps in f create kinks in F. F remains continuous but loses differentiability where f jumps.

7. Differentiability is the stronger property. It's rarer and more useful for calculus.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Your task (20 minutes):

Consider $f(x) = \begin{cases} x^2 & \text{if } x \leq 2 \\ 8 - x & \text{if } x > 2 \end{cases}$

1. Is f continuous at x = 2? Show all three conditions or explain why one fails.

2. Is f differentiable at x = 2? Compute the left and right derivatives.

3. Define $F(x) = \int_0^x f(t) \, dt$. Write a piecewise formula for F(x) on [0, 4].

4. Is F differentiable at x = 2? Why or why not?

Success looks like: You've written explicit limits, computed derivatives from first principles at the boundary point, shown that F'(x) = f(x) wherever f is continuous, and identified the exact location where F fails to be differentiable (if anywhere).

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