Textbook Question
a. Let f(x) be a function satisfying |f(x)| ≤ x² for −1 ≤ x ≤ 1. Show that f is differentiable at x = 0 and find f′(0).
2. Intro to Derivatives
Differentiability
4:55 minutesProblem 3.2.60b
Textbook Question
b. Show that
f(x) = { x² sin(1/x), x ≠ 0
0, x = 0
is differentiable at x = 0 and find f′(0).
Verified step by step guidance1
First, understand the definition of differentiability at a point. A function f(x) is differentiable at x = a if the limit of the difference quotient exists: lim(x -> a) [f(x) - f(a)] / (x - a).
For the given function f(x), we need to check differentiability at x = 0. The function is defined as f(x) = x² sin(1/x) for x ≠ 0 and f(x) = 0 for x = 0.
Calculate the difference quotient for f(x) at x = 0: [f(x) - f(0)] / (x - 0) = [x² sin(1/x) - 0] / x = x sin(1/x).
Evaluate the limit of x sin(1/x) as x approaches 0. Use the fact that |sin(1/x)| ≤ 1, so |x sin(1/x)| ≤ |x|. As x approaches 0, |x| approaches 0, thus the limit of x sin(1/x) is 0.
Since the limit exists and is equal to 0, f(x) is differentiable at x = 0. Therefore, f'(0) = 0.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Differentiability
A function is differentiable at a point if it has a defined derivative at that point. This means the function must be continuous at the point, and the limit of the difference quotient must exist as the point is approached. For f(x) to be differentiable at x = 0, we need to check these conditions specifically at x = 0.
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Limit Definition of Derivative
The derivative of a function at a point is defined as the limit of the difference quotient as the interval approaches zero: f'(a) = lim (h -> 0) [(f(a+h) - f(a))/h]. For f(x) at x = 0, this involves evaluating the limit of [x² sin(1/x)]/x as x approaches 0, which simplifies to finding the behavior of x sin(1/x) as x approaches 0.
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Squeeze Theorem
The Squeeze Theorem is used to find the limit of a function trapped between two other functions that have the same limit at a point. If -1 ≤ sin(1/x) ≤ 1, then -x ≤ x sin(1/x) ≤ x. As x approaches 0, both -x and x approach 0, allowing us to conclude that x sin(1/x) also approaches 0, which is crucial for proving differentiability at x = 0.
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