Continuity and Differentiability of Functions and Area Functions

TL;DR

Continuity means a function has no breaks or jumps, allowing you to draw it without lifting your pencil. Differentiability means a function is smooth enough to have a well-defined tangent line at every point. Area functions, like definite integrals, can also be continuous and differentiable, relating directly to the original function.

1. The Mental Model

Think of continuity as a bridge with no gaps: you can cross it smoothly. Differentiability is like that bridge also having no sharp corners or jagged edges, meaning you can always find a smooth slope at any point.

2. The Core Material

Continuity: No Breaks, No Jumps, No Holes

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A function $f(x)$ is continuous at a point $x=c$ if three conditions are met:
1. $f(c)$ exists (the function is defined at that point).
2. $\lim_{x \to c} f(x)$ exists (the limit from both sides approaches the same value).
3. $\lim_{x \to c} f(x) = f(c)$ (the limit equals the function's value).

If any of these fail, the function is discontinuous at $x=c$. Common discontinuities include:
* Removable (Holes): $\frac{(x-2)}{(x-2)(x+1)}$ has a hole at $x=2$.
* Jump: Piecewise functions often have these, like $f(x) = \begin{cases} x & x < 0 \\ x+1 & x \ge 0 \end{cases}$ at $x=0$.
* Infinite (Asymptotes): $1/x$ at $x=0$.

Differentiability: Smoothness and Tangent Lines

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A function $f(x)$ is differentiable at a point $x=c$ if its derivative $f'(c)$ exists. This means:
* The function is continuous at $x=c$ (differentiability implies continuity!).
* There's no sharp corner (like in $|x|$ at $x=0$).
* There's no cusp (a sharper corner).
* There's no vertical tangent line (like in $x^{1/3}$ at $x=0$).

The derivative at a point represents the instantaneous slope of the tangent line at that point.

graph TD
    A["Function is Differentiable at c"] --> B["Function is Continuous at c"]
    B --> C["f(c) exists"]
    B --> D["Limit of f(x) as x->c exists"]
    B --> E["Limit equals f(c)"]
    A --> F["No Sharp Corners/Cusps at c"]
    A --> G["No Vertical Tangent at c"]
    H["Function is NOT Differentiable at c"] --> I{"One or more conditions fail"}
    I --> J["Discontinuous at c"]
    I --> K["Sharp Corner/Cusp at c"]
    I --> L["Vertical Tangent at c"]

Area Functions (Definite Integrals)

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An area function, often denoted by $A(x)$, is defined as a definite integral with a variable upper limit:
$A(x) = \int_a^x f(t) \, dt$

Here, $f(t)$ is the integrand, and $a$ is a constant. This function $A(x)$ represents the net signed area under the curve of $f(t)$ from $a$ to $x$.

Continuity and Differentiability of Area Functions

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The Fundamental Theorem of Calculus, Part 1 (FTC1) is key here. It states that if $f(t)$ is a continuous function on an interval $[a, b]$, then the area function $A(x) = \int_a^x f(t) \, dt$ is differentiable on $(a,b)$, and its derivative is simply $f(x)$:
$A'(x) = \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(x)$

This means:
1. If $f(t)$ is continuous, then $A(x)$ is always differentiable.
2. Since differentiability implies continuity, $A(x)$ is also continuous if $f(t)$ is continuous.

So, if the integrand $f(t)$ is nice (continuous), then its area function $A(x)$ is even nicer (continuous and differentiable).

3. Worked Example

Let's consider the function $f(t) = t^2$. We want to define an area function $A(x) = \int_1^x t^2 \, dt$.

1. Is $f(t) = t^2$ continuous?
   Yes, $t^2$ is a polynomial, and all polynomials are continuous everywhere.

2. Based on FTC1, what can we say about $A(x)$?
   Since $f(t) = t^2$ is continuous, $A(x)$ must be differentiable, and therefore also continuous.

3. What is $A'(x)$?
   According to FTC1, $A'(x) = f(x)$. So, $A'(x) = x^2$.

4. Let's verify by actually calculating $A(x)$ and then differentiating it:
   $A(x) = \int_1^x t^2 \, dt = \left[ \frac{t^3}{3} \right]_1^x = \frac{x^3}{3} - \frac{1^3}{3} = \frac{x^3}{3} - \frac{1}{3}$

   Now, let's differentiate $A(x)$:
   $A'(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{1}{3} \right) = \frac{3x^2}{3} - 0 = x^2$.

   This matches exactly what FTC1 predicted!

4. Key Takeaways

• A function is continuous at a point if its value exists, its limit exists, and the value equals the limit.
• A function is differentiable at a point if it's continuous there and has no sharp corners, cusps, or vertical tangents.
• Differentiability implies continuity, but continuity does not imply differentiability.
• An area function $A(x) = \int_a^x f(t) \, dt$ represents the accumulated area under $f(t)$.
• If the integrand $f(t)$ is continuous, then its area function $A(x)$ is always differentiable.
• The derivative of an area function $\frac{d}{dx} \int_a^x f(t) \, dt$ is simply $f(x)$ (FTC1).

Common Mistakes to Avoid:
- Assuming continuity automatically means differentiability (e.g., $|x|$ is continuous but not differentiable at $x=0$).
- Forgetting to check all three conditions for continuity.
- Mixing up the variable of integration (t) with the upper limit variable (x) when applying FTC1.
- Not recognizing that a vertical tangent makes a function non-differentiable.

5. Now Try It

Consider the function $g(x) = \begin{cases} x^2 & x < 1 \\ x+1 & x \ge 1 \end{cases}$.
1. Determine if $g(x)$ is continuous at $x=1$.
2. Determine if $g(x)$ is differentiable at $x=1$.
3. Define an area function $B(x) = \int_0^x g(t) \, dt$. What can you say about the continuity and differentiability of $B(x)$ based on your findings for $g(x)$?

Success looks like clearly stating your reasoning for each part, showing how you checked the conditions for continuity and differentiability, and then applying FTC1 to discuss the area function.