Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist - AP Calculus AB Study Guide

The Core Idea: Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

This topic explores the fundamental relationship between two key concepts in calculus: continuity and differentiability. A function is continuous at a point if its graph is unbroken, meaning you can trace it without lifting your pencil. Differentiability is a more stringent condition; a function is differentiable at a point if it has a well-defined, non-vertical tangent line at that point. In essence, the graph must be not only connected but also "smooth" at that point.

The core principle is that for a function to be differentiable at a point, it must first be continuous at that point. However, the reverse is not always true; a function can be continuous at a point but fail to be differentiable there. This failure occurs at specific graphical features where a unique tangent line cannot be drawn: sharp corners, pointed cusps, or points with a vertical tangent line. Understanding this hierarchy—that differentiability implies continuity, but continuity does not guarantee differentiability—is crucial for analyzing the behavior of functions.

Key Theorems

The Differentiability Implies Continuity Theorem

This is the central theorem connecting the two concepts.

• Theorem: If a function $f$ is differentiable at $x=c$, then $f$ is continuous at $x=c$.

  • This means that the existence of a derivative $f^{\prime }(c)$ guarantees that the function is connected at that point. A derivative describes the slope of a tangent line, and you cannot have a well-defined tangent line at a point where the function has a gap or a hole.

• The Contrapositive: The logical equivalent of this theorem is often more useful for determining where a derivative does not exist.

  • Contrapositive Statement: If a function $f$ is not continuous at $x=c$, then $f$ is not differentiable at $x=c$.

  • This provides a preliminary test: if you identify a point of discontinuity (like a jump, hole, or vertical asymptote), you can immediately conclude that the function is not differentiable there.

Understanding Non-Differentiability

Even if a function is continuous at a point, it may still fail to be differentiable. This occurs when the graph is not "locally linear" or smooth. There are three specific graphical behaviors where a continuous function is not differentiable.

1. Corner: A corner occurs where the graph abruptly changes direction. At this point, the slope of the curve approaching from the left is different from the slope approaching from the right. Since the left-hand and right-hand limits of the slopes are not equal, a single, unique tangent line cannot be defined. The classic example is the absolute value function $f(x)=|x|$ at $x=0$.

2. Cusp: A cusp is a sharp point on a curve. It is a more extreme version of a corner where the slopes of the secant lines approach $\infty$ from one side and $-\infty$ from the other. The tangent lines become increasingly steep as you approach the cusp, ultimately becoming vertical from both sides. An example is the function $f(x)=x^{2/3}$ at $x=0$.

3. Vertical Tangent: A vertical tangent occurs at a point where the slopes of the secant lines approach either $\infty$ or $-\infty$ from both sides. The graph is smooth and continuous, but the tangent line at that point is perfectly vertical. Since a vertical line has an undefined slope, the derivative does not exist at this point. An example is the function $f(x)=\sqrt[3]{x}$ at $x=0$.

Core Concepts & Rules

• Differentiability implies Continuity: If a function has a derivative at a point, it must be continuous at that point.

• Continuity does NOT imply Differentiability: A function can be continuous at a point but not have a derivative there.

• Test for Non-Differentiability: A function $f(x)$ is not differentiable at $x=c$ if any of the following are true:

  1. The function is discontinuous at $x=c$.

  2. The graph of the function has a sharp corner at $x=c$.

  3. The graph of the function has a cusp at $x=c$.

  4. The graph of the function has a vertical tangent line at $x=c$.

Step-by-Step Example 1: Analyzing a Piecewise Function

Consider the piecewise function:

$$f(x)=\{\begin{array}{ll}{x}^2+1& \text{if }x\le 1\\ 3x-1& \text{if }x>1\end{array}$$

Determine if $f(x)$ is differentiable at $x=1$.

Step 1: Check for Continuity at $x=1$

For $f(x)$ to be continuous at $x=1$, the limit from the left, the limit from the right, and the function's value must all be equal.

• Function Value:$f(1)=(1{)}^2+1=2$.

• Left-Hand Limit:${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{-}}(x^2+1)=1^2+1=2$.

• Right-Hand Limit:${\lim }_{x\to 1^{+}}f(x)={\lim }_{x\to 1^{+}}(3x-1)=3(1)-1=2$.

Since ${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{+}}f(x)=f(1)=2$, the function is continuous at $x=1$. Because it is continuous, it might be differentiable. We must proceed to the next step.

Step 2: Check if the Derivatives from the Left and Right are Equal

For $f(x)$ to be differentiable at $x=1$, the derivative of the left-side piece must equal the derivative of the right-side piece at $x=1$.

• Find the derivative of each piece:

  • For $x<1$, $\frac{d}{dx}(x^2+1)=2x$.

  • For $x>1$, $\frac{d}{dx}(3x-1)=3$.

  So, $$f^{\prime }(x)=\{\begin{array}{ll}2x& \text{if }x<1\\ 3& \text{if }x>1\end{array}$$

• Evaluate the limit of the derivatives as $x\to 1$:

  • Derivative from the left:${\lim }_{x\to 1^{-}}{f}^{\prime }(x)={\lim }_{x\to 1^{-}}(2x)=2(1)=2$.

  • Derivative from the right:${\lim }_{x\to 1^{+}}{f}^{\prime }(x)={\lim }_{x\to 1^{+}}(3)=3$.

Step 3: Conclude

Since the derivative from the left (2) does not equal the derivative from the right (3), the derivative $f^{\prime }(1)$ does not exist. The graph of $f(x)$ has a corner at $x=1$. Therefore, $f(x)$ is continuous but not differentiable at $x=1$.

Step-by-Step Example 2: Analyzing a Graph

The graph of a function $f(x)$ is shown below. Identify all x-values on the interval $(a,k)$ where $f(x)$ is not differentiable, and state the reason for each.