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Chapter 6: Problem 11

Give an example of a uniformly continuous function on \([0,1]\) that is differentiable on \((0,1)\) but whose derivative is not bounded on \((0,1)\).

Short Answer

Expert verified
An example of such a function is given by \(f(x) = x^2 \sin\frac{1}{x}\) for \(x \neq 0\) and \(f(x) = 0\) for \(x = 0\). The function's derivative is given by \(f'(x) = 2x \sin\frac{1}{x} - \cos\frac{1}{x}\) for \(x \neq 0\) and \(f'(0) = 0 \), but it is not bounded within the interval \((0,1)\).

Step by step solution

01

Defining the function

Firstly, we must define a function that fulfills the required conditions. An example of such a function is \(f(x) = x^2 \sin\frac{1}{x}\) for \(x \neq 0\) and \(f(x) = 0\) for \(x = 0\). This function is differentiable on \((0,1)\), but its derivative is not bounded.
02

Calculating the derivative

Next, we need to calculate the derivative of our function using chain rule. The derivative \(f'(x)\) of the function is given by \(f'(x) = 2x \sin\frac{1}{x} - \cos\frac{1}{x} \), for \(x \neq 0\) and \(f'(0) = 0 \). As we can observe the derivative of \( f(x) \), although existing for \(x \in (0,1)\), is not bounded due to the changing values of cosine.

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