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Differentiability
& Continuity

Explore the deep connection between differentiability and continuity: when derivatives exist, when they don't, and why it matters.

AP Calculus AB

What Is Continuity?

A function \( f \) is continuous at \( x = a \) if three conditions hold simultaneously: the function is defined there, the limit exists, and they match up.

1
\( f(a) \) exists: the function is defined at the point.
2
\( \displaystyle\lim_{x \to a} f(x) \) exists: the left and right limits agree.
3
\( \displaystyle\lim_{x \to a} f(x) = f(a) \): the limit equals the function value.

What Is Differentiability?

A function \( f \) is differentiable at \( x = a \) if the derivative exists there. The derivative is defined as the limit:

This limit must exist and be finite. Geometrically, it means the graph has a unique, non-vertical tangent line at the point \( (a, f(a)) \).

Smooth vs Non-Smooth Graphs

Differentiable Tangent exists Not Differentiable Corner: no single tangent

The Core Theorem

In logical notation: Differentiability \( \Rightarrow \) Continuity. Differentiability is a stronger condition than continuity. Every differentiable function is automatically continuous.

The Implication Diagram

Continuous Functions Differentiable Functions
\(|x|\)
\(\sin x\)
\(\text{Differentiable} \subseteq \text{Continuous}\)

Continuity vs Differentiability

Continuity Differentiability
Definition \( \lim_{x \to a} f(x) = f(a) \) \( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \) exists
Geometric meaning No breaks or jumps Smooth, unique tangent line
Strength Weaker condition Stronger condition
Implication Does NOT imply differentiable Always implies continuous
Counterexample \( |x| \) at \( x=0 \) N/A: no counterexample exists

When Derivatives Do Not Exist

Even if \( f \) is continuous at \( x = a \), the derivative \( f'(a) \) fails to exist in four classic situations:

Corner
Left and right derivatives exist but are not equal. Example: \( |x| \) at \( x=0 \).
Cusp
One-sided derivatives approach \( +\infty \) and \( -\infty \). Example: \( x^{2/3} \) at \( x=0 \).
Vertical Tangent
Both one-sided derivatives approach \( +\infty \) or \( -\infty \). Example: \( x^{1/3} \) at \( x=0 \).
Discontinuity
If \( f \) is not continuous at \( a \), then \( f \) is automatically not differentiable there.

Four Failure Modes Visualized

Corner Cusp Vertical Tangent Discontinuity

Why Differentiability Implies Continuity

The proof is elegant and short. We want to show \( \lim_{x \to a} f(x) = f(a) \).

1
Rewrite the difference: \( f(x) - f(a) = \dfrac{f(x)-f(a)}{x-a} \cdot (x-a) \)
2
Take the limit: \( \displaystyle\lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \cdot \lim_{x \to a}(x-a) \)
3
Evaluate: \( = f'(a) \cdot 0 = 0 \). Therefore \( \lim_{x \to a} f(x) = f(a) \). ✓

How to Check Differentiability

1
Check continuity first. If \( f \) is not continuous at \( x = a \), stop: the function is not differentiable.
2
Compute the left-hand derivative: \( \displaystyle\lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h} \)
3
Compute the right-hand derivative: \( \displaystyle\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h} \)
4
Compare: If both one-sided limits exist, are finite, and are equal, then \( f'(a) \) exists.

Example 1: \( f(x) = |x| \) at \( x = 0 \)

x y
slope = \(-1\)
slope = \(+1\)
1
Continuity: \( \lim_{x \to 0} |x| = 0 = f(0) \). ✓ Continuous.
2
Left derivative: \( \lim_{h \to 0^-} \frac{|h|}{h} = \frac{-h}{h} = \) \( -1 \)
3
Right derivative: \( \lim_{h \to 0^+} \frac{|h|}{h} = \frac{h}{h} = \) \( +1 \)
4
Conclusion: \(-1\) \( \neq \) \(+1\), so \( f'(0) \) does not exist.

Example 2: Piecewise Function

1
Continuity: Left: \( 1^2 = 1 \). Right: \( 2(1)-1 = 1 \). \( f(1) = 1 \). ✓ Continuous.
2
Left derivative: \( \frac{d}{dx}[x^2] \big|_{x=1} = 2x\big|_{x=1} = \) \( 2 \)
3
Right derivative: \( \frac{d}{dx}[2x-1] \big|_{x=1} = \) \( 2 \)
4
Conclusion: Both equal \( 2 \), so \( f'(1) = 2 \). ✓ Differentiable.

Example 3: \( f(x) = x^{1/3} \) at \( x = 0 \)

x y Vertical tangent
1
Continuity: \( \lim_{x \to 0} x^{1/3} = 0 = f(0) \). ✓ Continuous.
2
Derivative attempt: \( \lim_{h \to 0} \frac{h^{1/3}}{h} = \lim_{h \to 0} h^{-2/3} = \) \( +\infty \)
3
Conclusion: The limit is infinite, so \( f'(0) \) does not exist. This is a vertical tangent.

Common Mistakes

✕ Converse Confusion
Assuming "continuous implies differentiable." The converse of the theorem is false: \( |x| \) is the classic counterexample.
✕ Forgetting Continuity Check
Jumping straight to one-sided derivatives without first verifying that \( f \) is continuous at the point.
✕ Ignoring Infinite Limits
If \( \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \pm\infty \), the derivative does not exist. A vertical tangent is not differentiable.

Summary Table

Situation at \( x = a \) Continuous? Differentiable?
Smooth curve Yes Yes
Corner (e.g., \( |x| \)) Yes No
Cusp (e.g., \( x^{2/3} \)) Yes No
Vertical tangent (e.g., \( x^{1/3} \)) Yes No
Jump discontinuity No No
Removable discontinuity No No

Question 1 of 5

If \( f \) is differentiable at \( x = 3 \), which of the following must be true?

Question 2 of 5

The function \( f(x) = |x - 2| \) at \( x = 2 \) is:

Question 3 of 5

Let \( g(x) = \begin{cases} x^2 + 1 & x \le 2 \\ 4x - 3 & x > 2 \end{cases} \). Is \( g \) differentiable at \( x = 2 \)?

Question 4 of 5

Which statement is always true?

Question 5 of 5

At \( x = 0 \), the function \( f(x) = x^{2/3} \) has a: