Differentiability and continuity; when derivatives fail to exist

Introduction: Smoothness and the Derivative

In our previous chapters, we learned how to calculate the derivative—the "instantaneous rate of change" or the slope of a curve at a single point. But does every function have a derivative at every point? The short answer is: No.

Think of a derivative as a measure of "smoothness." If a graph is smooth and connected, you can usually find its slope. If the graph has a "break," a "sharp point," or becomes "too steep," the derivative fails to exist. In this chapter, we will explore the critical link between continuity (the function being connected) and differentiability (the function having a derivative), and identify exactly where math "breaks" the derivative.

The Golden Rule: Differentiability Implies Continuity

There is a very important relationship you must memorize for the AP Exam: If a function is differentiable at \(x = c\), then it must be continuous at \(x = c\).

Imagine this as a one-way street. Being "differentiable" is a higher standard than being "continuous." If a function is smooth enough to have a derivative, it is guaranteed to be connected.
Symbolically: \(f'(c)\) exists \(\implies f(x)\) is continuous at \(x = c\).

The Trap: The reverse is not always true! Just because a function is continuous (connected) does not mean it is differentiable. The classic example is the absolute value function, \(f(x) = |x|\). It is one solid line, but it has a sharp point at the origin where the derivative does not exist.

When Derivatives Fail to Exist

On the AP Exam, you will often be asked to look at a graph or an equation and determine where the derivative fails to exist. There are four main "deal-breakers" for differentiability:

1. Any Discontinuity

If a function is not continuous at \(x = c\), it cannot be differentiable at \(x = c\). This includes:
• Removable Discontinuities (Holes): Even if the rest of the graph is smooth, a hole means there is no point to calculate a slope.
• Jump Discontinuities: The slope from the left and the right can't meet if the graph isn't even touching.
• Infinite Discontinuities (Vertical Asymptotes): The function doesn't exist at the asymptote, so the derivative can't either.

2. Sharp Corners (or Cusps)

A derivative represents a unique slope. At a sharp corner, the slope from the left side is different from the slope from the right side.
Example: In \(f(x) = |x|\), the slope to the left of zero is \(-1\), but the slope to the right is \(+1\). Because the slopes don't agree at the exact moment they meet, the derivative at \(x = 0\) is undefined.

3. Vertical Tangents

Sometimes a function is continuous and smooth, but it becomes "infinitely steep" for a split second.
Example: The cube root function, \(f(x) = x^{1/3}\). At \(x = 0\), the tangent line is a perfectly vertical line. Since the slope of a vertical line is undefined (\(m = \infty\)), the derivative does not exist there.

4. Wild Oscillations

This is rare on the AP exam but good to know. If a function oscillates (waves up and down) so infinitely fast that it never settles on a single slope as it approaches a point, the derivative fails.

The Formal Test: Left-Hand vs. Right-Hand Derivatives

Don't worry if this seems tricky at first! When dealing with piecewise functions, you can prove differentiability by checking if the "slopes match." For a function to be differentiable at \(x = c\):

1. The function must be continuous at \(x = c\). Check that \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\).
2. The limit of the derivative from the left must equal the limit of the derivative from the right:
\(\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)\)

Quick Review Box:
• Differentiable? Yes, it's continuous.
• Continuous? Maybe differentiable, maybe not (check for sharp corners!).
• Sharp Point? Continuous, but not differentiable.
• Vertical Tangent? Continuous, but not differentiable.

Common Mistakes to Avoid

Mistake 1: Forgetting to check continuity first.
Sometimes students check if the slopes of a piecewise function match, but they forget to check if the two pieces actually touch! If they don't touch (a jump discontinuity), the derivative fails even if the slopes happen to be the same.

Mistake 2: Assuming a derivative exists because the function is defined.
Just because \(f(c)\) exists doesn't mean \(f'(c)\) does. Always look for those sharp corners or vertical spots.

Step-By-Step: Checking Differentiability for Piecewise Functions

Let's say you have a function \(f(x)\) defined as \(x^2\) for \(x \le 1\) and \(2x - 1\) for \(x > 1\). Is it differentiable at \(x = 1\)?

Step 1: Check Continuity.
Plug \(x = 1\) into both parts.
Left part: \(1^2 = 1\).
Right part: \(2(1) - 1 = 1\).
They match! The function is continuous.

Step 2: Check the Slopes (Derivatives).
Find the derivative of each part.
Derivative of \(x^2\) is \(2x\). At \(x = 1\), the slope is \(2(1) = 2\).
Derivative of \(2x - 1\) is \(2\). At \(x = 1\), the slope is \(2\).
They match! Since the function is continuous and the slopes match, the function is differentiable at \(x = 1\).

Key Takeaways

• Continuity is a prerequisite for differentiability. If a graph is broken, it has no derivative at that point.
• "Smoothness" is key. Sharp turns (corners/cusps) and vertical tangents are the most common ways a continuous function fails to be differentiable.
• Use your calculator. On calculator-active sections, you can graph the derivative to see if it has a hole or a vertical asymptote at a certain point.
• Notation check: Remember that \(f'(c)\) is simply the limit of the difference quotient. If that limit doesn't exist, the derivative doesn't exist.

Did you know? The term "differentiable" comes from the word "difference," specifically the "difference quotient" \(\frac{\Delta y}{\Delta x}\) that we use to define the slope!