Problem 16 - Essential Calculus

Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0\). $$ f(x)=\frac{x^{2}-1}{x},[-1,1] $$

Step-by-Step Solution

Answer

Rolle's theorem cannot be applied to the function \(f(x)=\frac{x^{2}-1}{x}\) on the interval [-1, 1] as the function is not continuous and differentiable on the entire interval. Therefore, there are no values of c in this interval such that \(f'(c)=0\).

Step 1: Check the conditions of Rolle's Theorem

First, check if the function \(f(x)=\frac{x^{2}-1}{x},[-1,1]\) is continuous and differentiable in the interval and if f(-1) = f(1). The function is not defined at x=0. Hence, the function is not continuous or differentiable on the entire interval. So, Rolle's Theorem cannot be applied.

Step 2: Apply Rolle's Theorem on subintervals

Although Rolle’s Theorem cannot be applied to the whole interval [-1, 1], it can be applied to the subintervals [-1, 0) and (0, 1]. First, find the derivative of the function, which is \(f'(x)=\frac{-1}{x^{2}}\). Now, consider the subinterval [-1, 0). f(a) = f(-1) = 2 and f(b) = f(0) is not defined, hence Rolle's Theorem can't be applied to this subinterval. Now, consider the subinterval (0, 1]. f(a) = f(0) is not defined and f(b) = f(1) = 0, hence Rolle's Theorem can't be applied to this subinterval either.

Step 3: Conclusion

Rolle’s Theorem is not applicable to the given function on the interval [-1, 1] and no such c exists in this interval for which \(f'(c) = 0\).

Key Concepts

Function Continuity

In calculus, a function is said to be continuous on a closed interval \([a, b]\) if there are no "breaks" or "holes" in its graph over that interval. For Rolle's Theorem to be applicable, the function must be continuous over the entire interval \([a, b]\). This means that for every point in the interval, the function should have a defined value, and small changes in input should not result in sudden jumps or interruptions in the electronic function's output.

In the given exercise, the function \( f(x)=\frac{x^{2}-1}{x} \) is checked for continuity over the interval \([-1, 1]\). It is found to not be continuous because the function is not defined at \(x=0\) where there's a division by zero. Hence, there is a "hole" in the graph at this point. This discontinuity at \(x=0\) is why Rolle's Theorem cannot be applied across the whole interval.

• Ensure that the function behaves without interruptions over the given interval to apply the theorem.

Differentiability

Differentiability is another key concept required for applying Rolle's Theorem. A function is differentiable at a point if it possesses a derivative at that point, indicating a well-defined tangent to the curve at that point. Essentially, differentiability implies not just smoothness in the function but also no sharp corners or cusps.

For Rolle's Theorem, the function must be differentiable over the open interval \( (a, b) \). In our example, the derivative can be calculated as \( f'(x)=\frac{-1}{x^{2}} \). However, because the function is not defined at \(x=0\), it cannot be differentiable there. There is also an issue with differentiability at \(x=0\) since the function doesn't exist. Without differentiability over the necessary interval, Rolle's Theorem cannot be applied.

• Verify both continuity and differentiability to make use of Rolle’s Theorem.

Subinterval Analysis

When the entirety of an interval does not meet the requirements for Rolle's Theorem, analyzing subintervals can be a useful approach.

In the exercise, since the entire interval \( [-1, 1] \) is problematic due to discontinuity and non-differentiability at \(x=0\), subintervals \( [-1, 0) \) and \( (0, 1] \) are considered. For these subintervals, even if some conditions like continuity might be satisfied, both endpoints of the subintervals must equal each other (i.e., \( f(a) = f(b) \)) for Rolle's Theorem to apply.

On closer inspection, it was realized that neither subinterval completely satisfies all conditions of Rolle's Theorem, particularly because \( f(0) \) is not defined for either subinterval. Therefore, the subinterval analysis reveals that there’s no valid \(c\) such that \( f'(c)=0 \) within these subintervals, indicating that Rolle's Theorem cannot be applied in this case.

• Check for endpoints' equality in values across subintervals when trying subinterval analysis.

Calculus

The broad field of calculus provides tools to study change and motion. Rolle's Theorem is a valuable result in differential calculus, emphasizing the importance of function behavior (continuity and differentiability) over an interval.

It offers insights into understanding conditions under which a function reaches horizontal tangents — points where the derivative is zero. Rolle's Theorem is a special case of the Mean Value Theorem. It becomes applicable only when the function is confirmed to be both continuous and differentiable within the necessary intervals, with endpoints having equal function values.

In the context of the exercise, we're introduced to various features of function behavior that limit applications of Rolle's Theorem. This is crucial in calculus to understand and predict how a function behaves across its domain.

• Use calculus to deduce properties of functions and their graphs.