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Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

Samuel Baker

Samuel Baker

9 min read

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Study Guide Overview

This study guide covers the relationship between continuity and differentiability in AP Calculus AB/BC. It explains how to determine differentiability using limits and identifies common points of non-differentiability: discontinuities, corners, cusps, and vertical tangents. Practice problems and exam tips are included to reinforce these concepts.

#AP Calculus AB/BC: Differentiability and Continuity - The Ultimate Review 🚀

Hey there, future calculus conqueror! Let's make sure you're totally prepped for anything the AP exam throws your way. This guide is designed to be your go-to resource for understanding differentiability and continuity, especially when time is tight. Let’s dive in!

#🔗 2.4 Connecting Differentiability and Continuity

#↗️ Continuity and Differentiability

Most functions you'll see in AP Calc are both continuous and differentiable, but it's crucial to know when things go sideways. Remember, for a function to be differentiable at a point, it must be smooth and look like a straight line when you zoom in super close. This line doesn't have to be horizontal!

Key Concept

A function that is differentiable at a point must also be continuous at that point. However, the reverse is not necessarily true: a continuous function is not always differentiable. 💡

Check out the graph of cos⁡(x)\cos(x)cos(x) below. Notice how zooming in at (0,1)(0,1)(0,1) makes it look like a straight line. That’s differentiability in action!

![Screen Recording 2023-12-14 at 2.16.40 PM.gif](Screen Recording 2023-12-14 at 2.16.40 PM.gif)

Zooming into graph of cos⁡(x)\cos(x)cos(x)

Graph created with Desmos

For a function to be differentiable at a point 'a', the left-hand limit of the derivative must equal the right-hand limit of the derivative, and both must equal the value of the derivative at that point:

lim⁡x→a−f′(x)=lim⁡x→a+f′(x)=f′(a)\lim_{x\to a^{-}} f'(x) = \lim_{x\to a^{+}} f'(x) = f'(a)x→a−lim​f′(x)=x→a+lim​f′(x)=f′(a)

Exam Tip

Remember: Differentiability implies continuity, but continuity does not imply differentiability. This is a common point of confusion! 🧐

#🚧 When is a Function NOT Differentiable?

A function is NOT differentiable if it has any of the following:

  • Discontinuities (jumps, holes, or vertical asymptotes)
  • Sharp Turns (corners or cusps)
  • Vertical Tangents

Let's break down each scenario:

#🕳 Discontinuous Graphs

#Unequal Derivatives

If a graph has a discontinuity, the derivatives from the left and right sides will not match. Take a look at g(x)=1x+2g(x) = \frac{1}{x+2}g(x)=x+21​.

![Screen Shot 2023-11-29 at 9.36.07 PM.png](Screen Shot 2023-11-29 at 9.36.07 PM.png)

Graph of g(x)=1x+2g(x) = \frac{1}{x+2}g(x)=x+21​

Graph created with Desmos

Notice that lim⁡x→−2−g′(x)<0\lim_{x\to -2^{-}} g'(x) < 0limx→−2−​g′(x)<0 and lim⁡x→−2+g′(x)>0\lim_{x\to -2^{+}} g'(x) > 0limx→−2+​g′(x)>0. Since the derivative values are different, g(x)g(x)g(x) is not differentiable at x=−2x = -2x=−2.

#Jump Discontinuity

Similarly, with a jump discontinuity, the derivatives from the left and right sides will not be equal. Observe the graph below:

![Screen Shot 2023-12-14 at 6.43.17 PM.png](Screen Shot 2023-12-14 at 6.43.17 PM.png)

Graph of jump discontinuity

Graph created with Desmos

Here, lim⁡x→2−f′(x)>0\lim_{x\to 2^{-}} f'(x) > 0limx→2−​f′(x)>0 and lim⁡x→2+f′(x)<0\lim_{x\to 2^{+}} f'(x) < 0limx→2+​f′(x)<0, so the function is not differentiable at x=2x = 2x=2.

#Removable Discontinuity

Even with a removable discontinuity, like in h(x)=x2−4x−2h(x) = \frac{x^2 - 4}{x - 2}h(x)=x−2x2−4​, the function is not differentiable at the point of discontinuity. Even though the left and right limits of the derivative are equal, h′(2)h'(2)h′(2) does not exist.

![Screen Shot 2023-12-14 at 6.46.09 PM.png](Screen Shot 2023-12-14 at 6.46.09 PM.png)

Graph of removable discontinuity

Graph created with Desmos

Common Mistake

A common mistake is to assume that if the left and right limits of the derivative are equal, the function is differentiable at that point. Remember to also check if f′(a)f'(a)f′(a) exists at that point. 🧐

#⬆️ Vertical Tangents

Vertical tangents occur when the derivative approaches infinity (or negative infinity). For example, look at f(x)=2(x+2)13f(x) = 2(x+2)^{\frac{1}{3}}f(x)=2(x+2)31​.

![Screen Shot 2023-12-14 at 7.05.11 PM.png](Screen Shot 2023-12-14 at 7.05.11 PM.png)

Graph with vertical tangent

Graph created with Desmos

At x=−2x = -2x=−2, the tangent line is vertical, meaning the derivative is undefined. Therefore, the function is not differentiable at x=−2x = -2x=−2.

#📐 Corners and Cusps

Corners and cusps are points where the derivative changes abruptly, making the function non-differentiable.

![Screen Shot 2023-12-14 at 7.04.13 PM.png](Screen Shot 2023-12-14 at 7.04.13 PM.png)

Graph with a vertical tangent

Graph created with Desmos

For example, in f(x)=2(x+2)13f(x) = 2(x+2)^{\frac{1}{3}}f(x)=2(x+2)31​, there is a cusp at x=−2x=-2x=−2. Also, in g(x)=∣x∣g(x) = |x|g(x)=∣x∣, there's a sharp corner at x=0x=0x=0.

![Screen Shot 2023-12-14 at 8.34.13 PM.png](Screen Shot 2023-12-14 at 8.34.13 PM.png)

Graph of piecewise function with a corner

Graph created with Desmos

Memory Aid

Think of it this way: If you can't smoothly trace the curve with your pencil without lifting it or changing direction abruptly, it's not differentiable at those points. ✏️

#🧮 Practice Problems

Let's solidify your understanding with some practice!

#1) Graphically Identifying Differentiability at a Point

How many points on the graph of f(x)f(x)f(x) below are not differentiable?

![Screen Shot 2023-12-14 at 10.00.20 PM.png](Screen Shot 2023-12-14 at 10.00.20 PM.png)

Graph of piecewise function with a number of points that are not differentiable

Graph created with Desmos

Answer: There are 6 points where the function is not differentiable:

  1. Infinite discontinuity at x=−1x = -1x=−1
  2. Jump discontinuity at x=2x = 2x=2
  3. Cusp at (1,2)(1, 2)(1,2)
  4. Cusp at (9,3)(9, 3)(9,3)
  5. Corner at (6,0)(6, 0)(6,0)
  6. Vertical tangent at (10,4)(10, 4)(10,4)
Practice Question

Multiple Choice Questions:

  1. At which of the following values of x is the function f(x)=∣x−3∣f(x) = |x-3|f(x)=∣x−3∣ not differentiable? (A) -3 (B) 0 (C) 3 (D) All real numbers

  2. Which of the following statements is true about a function that is continuous at x=c? (A) The function must be differentiable at x=c. (B) The function must have a limit at x=c. (C) The function must be defined at x=c. (D) The function must be both differentiable and have a limit at x=c.

Free Response Question:

Consider the function f(x)f(x)f(x) defined as follows:

f(x)={x2+1,x≤13x−1,x>1f(x) = \begin{cases} x^2 + 1, & x \leq 1 \\ 3x - 1, & x > 1 \end{cases}f(x)={x2+1,3x−1,​x≤1x>1​

(a) Is f(x)f(x)f(x) continuous at x=1x=1x=1? Justify your answer. (b) Is f(x)f(x)f(x) differentiable at x=1x=1x=1? Show your work.

Scoring Rubric:

(a) Continuity at x=1 (3 points)

  • 1 point: Checks the left-hand limit
  • 1 point: Checks the right-hand limit
  • 1 point: Concludes continuity with justification

(b) Differentiability at x=1 (6 points)

  • 2 points: Finds the derivative of x2+1x^2 + 1x2+1
  • 1 point: Evaluates the left-hand derivative at x=1
  • 2 points: Finds the derivative of 3x−13x-13x−1
  • 1 point: Evaluates the right-hand derivative at x=1 and concludes differentiability (or lack thereof) with justification.

#2) Determining Differentiability at a Point

Let's tackle a free-response question (FRQ) from the 2003 AP Calculus AB exam. The function f(x)f(x)f(x) is defined as:

f(x)={85x+1,x<325x+2,x≥3f(x) = \begin{cases} \frac{8}{5}\sqrt{x+1}, & x < 3 \\ \frac{2}{5}x + 2, & x \geq 3 \end{cases}f(x)={58​x+1​,52​x+2,​x<3x≥3​

Given that f(x)f(x)f(x) is continuous at x=3x = 3x=3, is it also differentiable?

#👉 Checking the Left-hand Side

Find the derivative of the left-hand side:

f′(x)=ddx(85x+1)=45x+1f'(x) = \frac{d}{dx} \left( \frac{8}{5}\sqrt{x+1} \right) = \frac{4}{5\sqrt{x+1}}f′(x)=dxd​(58​x+1​)=5x+1​4​

Then, find the limit as xxx approaches 3 from the left:

lim⁡x→3−f′(x)=453+1=45(2)=25\lim_{x\to 3^{-}} f'(x) = \frac{4}{5\sqrt{3+1}} = \frac{4}{5(2)} = \frac{2}{5}x→3−lim​f′(x)=53+1​4​=5(2)4​=52​

#👉 Checking the Right-hand Side

Find the derivative of the right-hand side:

f′(x)=ddx(25x+2)=25f'(x) = \frac{d}{dx} \left( \frac{2}{5}x + 2 \right) = \frac{2}{5}f′(x)=dxd​(52​x+2)=52​

Then, find the limit as xxx approaches 3 from the right:

lim⁡x→3+f′(x)=25\lim_{x\to 3^{+}} f'(x) = \frac{2}{5}x→3+lim​f′(x)=52​

#👆 Checking f′(x)f'(x)f′(x) exists at x=3x = 3x=3

Since the derivative of the right-hand side is 25\frac{2}{5}52​, the derivative at x=3x=3x=3 is also 25\frac{2}{5}52​.

Since lim⁡x→3−f′(x)=lim⁡x→3+f′(x)=f′(3)=25\lim_{x\to 3^{-}} f'(x) = \lim_{x\to 3^{+}} f'(x) = f'(3) = \frac{2}{5}limx→3−​f′(x)=limx→3+​f′(x)=f′(3)=52​, the function f(x)f(x)f(x) is differentiable at x=3x = 3x=3!

![Screen Shot 2023-12-14 at 11.39.18 PM.png](Screen Shot 2023-12-14 at 11.39.18 PM.png)

Graph of piecewise function

Graph created with Desmos

#🌟 Final Exam Focus

High-Priority Topics:

  • Understanding the relationship between continuity and differentiability
  • Identifying points of non-differentiability (discontinuities, corners, cusps, vertical tangents)
  • Using limits to determine differentiability
Exam Tip

Exam Tips:

  • Always check for continuity before checking for differentiability.
  • Pay close attention to piecewise functions and their behavior at the transition points.
  • Practice identifying non-differentiable points graphically and algebraically.
Quick Fact

Quick Facts:

  • A function must be continuous to be differentiable.
  • Corners, cusps, and vertical tangents are always points of non-differentiability.
Common Mistake

Common Pitfalls:

  • Assuming continuity implies differentiability.
  • Forgetting to check if f′(a)f'(a)f′(a) exists when evaluating differentiability at a point.
  • Misinterpreting the meaning of vertical tangents.

#🎉 Closing

You’ve got this! With a solid grasp of these concepts and a bit more practice, you'll be ready to tackle any differentiability and continuity question that comes your way. Keep up the great work, and remember, you're doing amazing! 🚀

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Question 1 of 10

If a function is differentiable at a point, what must also be true at that point? 🤔

It must be discontinuous

It must be continuous

It must have a vertical tangent

It must have a sharp corner