Question

\(\mathrm{f}(\mathrm{x})=\mathrm{x} \cdot \sin (1 / \mathrm{x})\) and \(\mathrm{x} \in[-1,1]\). Also \(\mathrm{f}(0)=0\) then. (a) \(\mathrm{f}(\mathrm{x})\) is continuous in \([-1,1]\) (b) Roll's theorem is applicable in \([-1,1]\) (c) First mean value theorem is applicable in \([-1,1]\) (d) none of these

Short answer

The short answer is: \(f(x)\) is continuous in \([-1, 1]\) and First Mean Value Theorem is applicable in \([-1, 1]\).

Step-by-step solution

Function Continuity

A function is continuous at a point if the limit exists at that point, and the limit is equal to the function's value at that point. We need to check the continuity of the function \(f(x) = x \cdot \sin(1/x)\) over the interval \([-1, 1]\). If \(\mathrm{x}\neq 0\), the function is continuous because the product of two continuous functions (x and \(\sin(1/x)\)) is continuous. To check the continuity at x=0, we need to find the limit as x approaches 0. Let's calculate the limit: \(\lim_{x\to 0} x \cdot \sin(1/x)\) divide by x, the limit becomes: \(\lim_{x\to 0} \sin(1/x)\) Since sine function has output value in interval \([-1, 1]\), the limit as x approaches 0 for \(\sin(1/x)\) lies between -1 and 1. Hence, \(\lim_{x\to 0} x \cdot \sin(1/x) = 0\), and since f(0) = 0, the function is continuous at x = 0. Therefore, f(x) is continuous in the interval [-1, 1].

Function Differentiability

To check differentiability of the function, we need to find its derivative, and see if it exists in the given interval. The derivative of f(x) when x ≠ 0: \(f'(x) = \sin(1/x) + x \cdot (-\frac{1}{x^2})\cos(1/x)\) We cannot be sure about differentiability at x = 0, so let's check it separately. As x approaches 0, we find the limit: \(\lim_{x\to 0} f'(x)\) Since it is difficult to find the limit directly, we will use the Squeeze Theorem. Firstly, we have: -1 ≤ \(\sin(1/x)\) ≤ 1 Now, we add x\(\cdot\)(-1/x²)\(\cos(1/x)\) to each term: -1 + x\(\cdot\)(-1/x²)\(\cos(1/x)\) ≤ \(\sin(1/x) + x\cdot(-1/x²)\cos(1/x)\) ≤ 1 + x\(\cdot\)(-1/x²)\(\cos(1/x)\) Since the left part approaches 0 as x approaches 0, and the right part also approaches 0 as x approaches 0, we can conclude that the middle part approaches 0 as well, according to the Squeeze theorem. Therefore, the function is differentiable at x = 0. Now, we can conclude that f(x) is differentiable in the interval [-1, 1].

Conclusions

a) The function f(x) is continuous in [-1, 1]. b) Roll's theorem states that if a function is continuous on [a, b] and differentiable on (a, b), and its values at a and b are equal ( f(a) = f(b) ), then there exists at least one point c in the interval (a, b) such that its derivative at c is 0 ( f'(c) = 0 ). Since f is continuous, differentiable but not equal in the given interval [-1, 1] i.e. f(-1) ≠ f(1), Roll's theorem is not applicable in [-1, 1]. c) First mean value theorem states that if a function is continuous on [a, b] and differentiable on (a, b), then there exists a point c in the interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). Since f is continuous and differentiable in the given interval [-1, 1], the First Mean Value Theorem is applicable in [-1, 1]. d) Since the function f(x) is continuous, differentiable, and the First Mean Value Theorem is applicable in [-1, 1], "none of these" is incorrect. So, the correct answers are: - f(x) is continuous in [-1, 1]. - First Mean Value Theorem is applicable in [-1, 1].

Key concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus. It connects the behavior of a function on a closed interval with its derivatives. Imagine you're driving a car, and you know your starting and ending velocities over a journey; the MVT assures you that your speed matched your average speed at least once during the trip.

Here's how it works for a function:

• A function must be continuous on a closed interval \([a, b]\).
• It must also be differentiable on the open interval \( (a, b) \).

When these conditions are met, there exists at least one point \(c\) in \( (a, b) \) such that:
\[f'(c) = \frac{f(b) - f(a)}{b - a}\]

For our function \(f(x) = x \cdot \sin(1/x)\), it's continuous and differentiable in \([-1, 1]\), making the MVT applicable. This means there's some point where the instantaneous rate of change (or slope of the tangent) equals the average rate of change across the interval.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is all about continuity and connecting points on a graph. It tells us that for any value between \(f(a)\) and \(f(b)\) in a continuous function, there's always a corresponding point where the function takes that value.

Here's the breakdown:

• The function has to be continuous on a closed interval \([a, b]\).
• If a value \(N\) exists between \(f(a)\) and \(f(b)\), then there is at least one \(c\) in \( (a, b) \) such that \(f(c) = N\).

For our function \(f(x) = x \cdot \sin(1/x)\), this theorem supports that it can reach any value between \(f(-1)\) and \(f(1)\), as long as those values lie between the calculated endpoints on the interval.

Squeeze Theorem

The Squeeze Theorem is like having two friends holding you up when you're walking on a narrow path. It helps find limits when a function is "squeezed" between two other functions whose limits are well-known.

Here's how the theorem is structured:

• If \(g(x) \leq f(x) \leq h(x)\) for all \(x\) in some interval around \(a\), except possibly at \(a\).
• If \(\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} f(x) = L\).

The Squeeze Theorem was used in our original solution to determine the continuity and differentiability of the function \(f(x) = x \cdot \sin(1/x)\) at \(x = 0\). By squeezing it between two simpler functions, we proved that the limit is indeed 0, thus confirming continuity and differentiability.

Functions

Functions are the core building blocks of calculus and mathematics in general. Think of a function like a machine that takes input values, processes them, and yields an output.

Understanding functions includes:

• The domain: the set of allowable input values.
• The range: the set of possible outputs.
• The behavior: how it changes over an interval or point.

In the exercise, our function \(f(x) = x \cdot \sin(1/x)\) has some intriguing characteristics. It's continuous across \([-1, 1]\), including at \(x = 0\), where the behavior is governed by the sine function's oscillations. Understanding these attributes allows us to apply theorems like MVT and more to analyze and understand the function deeply.