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 (integrals) are always continuous, and they're differentiable precisely where their integrand is continuous—this link between integration and differentiation is the Fundamental Theorem of Calculus.

1. The Mental Model

Think of continuity as "you can draw it without lifting your pen," and differentiability as "you can draw it with no sharp corners or vertical tangents." A differentiable function is always continuous, but continuity alone isn't enough—a smooth path can have a kink. Area functions sit at the heart of calculus: as you slide the right endpoint of an integral along the x-axis, the area under the curve changes, and that rate of change is the original function itself.

2. The Core Material

Continuity: The Definition

A function $f$ is continuous at $x = a$ if three things hold:
1. $f(a)$ exists (the function is defined at $a$)
2. $\lim_{x \to a} f(x)$ exists
3. $\lim_{x \to a} f(x) = f(a)$

In plain terms: the function value matches what the function is "heading toward" from both sides. If any piece fails, the function jumps, has a hole, or shoots to infinity—it's not continuous.

Example: $f(x) = x^2$ is continuous everywhere because for any $a$, $\lim_{x \to a} x^2 = a^2 = f(a)$. But $g(x) = \frac{1}{x}$ is not continuous at $x = 0$ because $g(0)$ doesn't exist.

Differentiability: Slope Everywhere

A function is differentiable at $x = a$ if the derivative exists there:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

This limit must exist and be finite. Geometrically, the graph must have a well-defined tangent line at that point—no corners, cusps, or vertical tangents allowed.

Key fact: If $f$ is differentiable at $a$, then $f$ is also continuous at $a$. But continuity does not guarantee differentiability. For example, $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there (it has a corner).

graph TD
    A["Is the function defined at x = a?"] -->|No| B["Not continuous at a"]
    A -->|Yes| C["Does the limit as x→a exist?"]
    C -->|No| B
    C -->|Yes| D["Does limit equal f(a)?"]
    D -->|No| B
    D -->|Yes| E["Continuous at a"]
    E -->|Does the derivative exist?| F["Differentiable at a"]
    E -->|No derivative| G["Continuous but not differentiable"]

Area Functions and the Fundamental Theorem

Define an area function:
$$F(x) = \int_a^x f(t) \, dt$$

This represents the area under the curve $y = f(t)$ from a fixed point $a$ to a variable point $x$. The big reveal is in the Fundamental Theorem of Calculus:

If $f$ is continuous on $[a, b]$, then $F(x) = \int_a^x f(t) \, dt$ is differentiable on $(a, b)$, and $F'(x) = f(x)$.

This means: the rate at which area accumulates equals the height of the function at that point.

Why? When you move $x$ to $x + h$, the area increases by roughly $f(x) \cdot h$ (a thin rectangle). So $\frac{\Delta F}{h} \approx f(x)$, and in the limit, $F'(x) = f(x)$.

Area functions are always continuous (even if $f$ has jumps), because the integral "smooths out" the integrand. But they're differentiable only where $f$ itself is continuous. If $f$ jumps, $F$ is continuous but has a corner there.

3. Worked Example

Problem: Let $f(t) = 2t$ and define $F(x) = \int_0^x 2t \, dt$. Find $F(x)$, verify it's continuous at $x = 1$, and check that $F'(1) = f(1)$.

Step 1: Evaluate the integral.
$$F(x) = \int_0^x 2t \, dt = [t^2]_0^x = x^2$$

Step 2: Check continuity at $x = 1$.
- $F(1) = 1^2 = 1$ ✓
- $\lim_{x \to 1} x^2 = 1$ ✓
- They match, so $F$ is continuous at $x = 1$.

Step 3: Compute $F'(1)$.
$$F'(x) = \frac{d}{dx}(x^2) = 2x$$
$$F'(1) = 2(1) = 2$$

And $f(1) = 2(1) = 2$, so indeed $F'(1) = f(1)$. ✓

The Fundamental Theorem worked: differentiation reversed integration.

4. Key Takeaways

4.1 Most Important Concepts

1. Continuity at a point means $f(a)$, the limit as $x \to a$, and the function value all agree—no breaks or holes.

2. Differentiability is stronger than continuity: it requires a well-defined slope, so no corners or vertical tangents allowed.

3. Area functions are always continuous, even when their integrand has jumps or is discontinuous.

4. The Fundamental Theorem links integration and differentiation: if $F(x) = \int_a^x f(t) \, dt$ and $f$ is continuous, then $F'(x) = f(x)$.

5. Differentiability implies continuity, but not vice versa; continuity is the weaker condition.

6. Piecewise functions can be continuous (if the pieces match at the boundary) and may or may not be differentiable (check if the left and right derivatives agree at the break point).

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Exercise: Consider the piecewise function
$$f(x) = \begin{cases} x & \text{if } x < 1 \\ 2 & \text{if } x \geq 1 \end{cases}$$

(a) Is $f$ continuous at $x = 1$? Justify by checking the three conditions.

(b) Is $f$ differentiable at $x = 1$? Compute the left derivative $\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$ and the right derivative $\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$ and compare.

(c) Now define $F(x) = \int_0^x f(t) \, dt$. Find a formula for $F(x)$ by splitting the integral into two regions: $[0, 1)$ and $[1, x]$. (Assume $x \geq 1$.)

(d) Is $F$ continuous at $x = 1$? Is $F$ differentiable at $x = 1$?

Success looks like: You've shown that $f$ is continuous but not differentiable at 1 (different left and right derivatives), computed $F$ as a piecewise formula, and explained that $F$ is continuous everywhere but has a corner at $x = 1$ where $F'$ jumps from 1 to 2.

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