Differentiability at a point: algebraic (function isn't differentiable) | Khan Academy
TLDRThe video script discusses the continuity and differentiability of a piecewise function 'g' at x=1. It establishes that 'g' is continuous at x=1 by evaluating the left and right-hand limits, which both equal zero, matching g(1). However, differentiability is questioned by examining the limit of (g(x) - g(1))/(x - 1) from both sides. The left-hand limit yields a slope of one, while the right-hand limit indicates a zero slope, revealing that the function is continuous but not differentiable at x=1.
Takeaways
- 📚 The function in question is defined piecewise and needs to be evaluated for continuity and differentiability at x=1.
- 🔍 To determine continuity, the value of g(1) must match the limit of g(x) as x approaches 1 from both the left and right sides.
- 👉 For continuity, g(1) is calculated as 0 based on the given piecewise definition, which falls under the first case of the function.
- 👈 The left-hand limit of g(x) as x approaches 1 is also 0, which is found by substituting x=1 into the left-hand side of the piecewise function.
- 👉 The right-hand limit of g(x) as x approaches 1 is similarly 0, confirming that the function is continuous at x=1.
- 🚫 The script eliminates options suggesting the function is not continuous, based on the established continuity.
- 🤔 Differentiability requires a defined limit for the derivative as x approaches 1, which is g(x) - g(1) / (x - 1).
- 🔄 The left-hand limit of the derivative is found to be 1, as g(x) simplifies to x - 1 for x < 1, and (x - 1) / (x - 1) equals 1.
- 🔄 The right-hand limit of the derivative is 0, as g(x) simplifies to (x - 1)^2 for x ≥ 1, and (x - 1)^2 / (x - 1) equals x - 1, which is 0 when x=1.
- 🚫 The function is not differentiable at x=1 because the left-hand and right-hand limits of the derivative are not equal.
- 📉 The final conclusion is that the function is continuous but not differentiable at x=1, based on the differing slopes approaching the point from either side.
Q & A
What is the main topic discussed in the video script?
-The main topic discussed in the video script is the continuity and differentiability of a piecewise function at x equals one.
What are the four possible outcomes for the function g at x equals one?
-The four possible outcomes for the function g at x equals one are: continuous but not differentiable, differentiable but not continuous, both continuous and differentiable, and neither continuous nor differentiable.
What is the first step in determining if the function g is continuous at x equals one?
-The first step in determining if the function g is continuous at x equals one is to check if g(1) is equal to the limit of g(x) as x approaches one from both the left and right sides.
What is the value of g(1) in the given piecewise function?
-The value of g(1) in the given piecewise function is zero, as it falls into the case where the function is defined as 1 - 1^2, which simplifies to 0.
How does the script evaluate the left-hand limit of g(x) as x approaches one?
-The script evaluates the left-hand limit by considering the definition of g(x) for x less than one, which is a continuous function for all real numbers, and substituting x with one, resulting in zero.
How does the script evaluate the right-hand limit of g(x) as x approaches one?
-The script evaluates the right-hand limit by considering the definition of g(x) for x greater than or equal to one, which is x - 1^2, and substituting x with one, resulting in zero.
Why can we conclude that the function g is continuous at x equals one?
-We can conclude that the function g is continuous at x equals one because both the left-hand limit and the right-hand limit of g(x) as x approaches one are equal to g(1), which is zero.
What is the criterion for differentiability of a function at a point?
-The criterion for differentiability of a function at a point is the existence of a defined limit as x approaches that point for the expression (g(x) - g(a)) / (x - a), where a is the point in question.
How does the script determine the differentiability of g(x) at x equals one?
-The script determines the differentiability of g(x) at x equals one by evaluating the limit of (g(x) - g(1)) / (x - 1) from both the left and right sides and checking if these limits are the same.
What does the script conclude about the differentiability of the function g at x equals one?
-The script concludes that the function g is not differentiable at x equals one because the limits from the left and right sides are not the same, indicating a corner or a cusp in the graph at that point.
What does the script suggest about the graph of the function g at x equals one?
-The script suggests that the graph of the function g is continuous at x equals one but has a corner or a cusp at that point, indicating a change in slope from one side to the other, which means it is not differentiable there.
Outlines
🔍 Continuity and Differentiability Analysis
The script begins with a question about the continuity and differentiability of a piecewise function 'g' at x=1. It explains the concept of continuity, stating that for a function to be continuous at a point, the value of the function at that point must equal the limit of the function as it approaches that point. The function 'g' is evaluated at x=1 and its left and right-handed limits are calculated, showing that it is continuous at x=1. The script then moves on to discuss differentiability, setting up the limit for the derivative from both sides of x=1, but the explanation is cut off before completion.
📉 Derivative Limit Analysis and Function Behavior
Continuing from the previous paragraph, the script discusses the differentiability of the function 'g' by evaluating the limit of (g(x) - g(1)) / (x - 1) from both the left and right sides of x=1. It simplifies the expression for the left-hand limit, finding it to be 1, and then evaluates the right-hand limit, which simplifies to x - 1, and ultimately to 0 when x=1. The script concludes that the function is continuous but not differentiable at x=1, as the left and right-hand limits of the derivative do not match, indicating a sharp point or 'corner' in the graph of the function at that point.
Mindmap
Keywords
💡Continuous
💡Differentiable
💡Limit
💡Piecewise function
💡Function g(x)
💡Left-hand limit
💡Right-hand limit
💡Derivative
💡Slope
💡Graph
Highlights
The function g is defined piecewise and its continuity and differentiability at x equals one are discussed.
For g to be continuous at x=1, g(1) must equal the limit of g(x) as x approaches 1.
g(1) is calculated to be 0, indicating a potential for continuity.
Left and right-hand limits of g(x) as x approaches 1 are both evaluated to be 0, confirming continuity.
Differentiability requires a defined limit for (g(x) - g(1)) / (x - 1) as x approaches 1.
The left-hand limit of g(x) / (x - 1) as x approaches 1 from the left is found to be 1.
The right-hand limit of g(x) / (x - 1) as x approaches 1 from the right is 0.
Different limits from the left and right sides indicate that g is not differentiable at x=1.
The function g is confirmed to be continuous but not differentiable at x=1.
The function's graph is continuous, but the slopes on either side of x=1 differ, indicating non-differentiability.
The left-hand limit approach simplifies g(x) to x - 1, showing a slope of 1.
The right-hand limit approach simplifies g(x) to x - 1 squared / (x - 1), resulting in a slope of 0.
The function's definition for x < 1 and x ≥ 1 is crucial for evaluating continuity and differentiability.
The video encourages viewers to pause and determine the function's properties themselves, fostering active learning.
The explanation methodically breaks down the process of evaluating continuity and differentiability step by step.
The function's behavior at x=1 is analyzed by comparing the function's value and its limit as x approaches 1.
The transcript uses a clear and educational approach to explain mathematical concepts, suitable for learning.
The conclusion that the function is continuous but not differentiable is reached through a logical and detailed analysis.
Transcripts
Browse More Related Video
2011 Calculus AB free response #6a | AP Calculus AB | Khan Academy
Justification with the intermediate value theorem: equation | AP Calculus AB | Khan Academy
Worked example: Continuity at a point | Limits and continuity | AP Calculus AB | Khan Academy
2019 AP Calculus AB Free Response Question #6
2011 AP Calculus AB Free Response #6
Introduction to limits | Limits | Differential Calculus | Khan Academy
5.0 / 5 (0 votes)
Thanks for rating: