Problem 15 - University Calculus: Early Transcendentals

Question

The function $$f(x)=\left\\{\begin{array}{ll} x, & 0 \leq x<1 \\ 0, & x=1 \end{array}\right.$$ is zero at \(x=0\) and \(x=1\) and differentiable on \((0,1),\) but its derivative on (0,1) is never zero. How can this be? Doesn't Rolle's Theorem say the derivative has to be zero somewhere in (0,1)\(?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Rolle's Theorem doesn't apply because the function isn't continuous on \([0,1]\).

Step 1: Understanding Rolle's Theorem

Rolle's Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there is at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

Step 2: Identifying the Interval and Conditions

The given function is defined as \( f(x) = x \) for \( 0 \leq x < 1 \) and \( f(x) = 0 \) for \( x = 1 \). It is differentiable on \((0,1)\) and evaluates to zero at \( x = 0 \) and \( x = 1 \).

Step 3: Checking Continuity on [0,1]

For Rolle's Theorem to apply, \( f(x) \) needs to be continuous on \([0,1]\), but it is not continuous since at \( x = 1 \), \( f(x) \) suddenly jumps from 1 to 0. Hence, the function is discontinuous at \( x = 1 \).

Step 4: Conclusion on Rolle's Theorem Applicability

Since \( f(x) \) is not continuous on the closed interval due to the discontinuity at \( x = 1 \), Rolle's Theorem cannot be applied. This allows for the derivative to never be zero on \((0,1)\) without contradicting Rolle's Theorem.

Key Concepts

Continuity and Its Importance

Continuity is a fundamental concept in calculus that describes a function's behavior. A function is continuous on an interval if, roughly speaking, you can draw it without lifting your pencil. This means there's no sudden jumps, breaks, or holes in the function's graph within that interval.
For the application of Rolle's Theorem, continuity is crucial because it ensures that the function behaves smoothly across the entire closed interval \([a, b]\).
In the provided exercise, the function \(f(x)\) is indeed smooth on the interval \((0, 1)\). However, it exhibits a discontinuity at \(x = 1\) due to the abrupt change from \(f(x) = x\) to \(f(x) = 0\). This discontinuity breaks the continuous path required by Rolle's Theorem.
To clearly illustrate continuity:

• No jumps: The function must not abruptly change in value.
• No breaks: The function must be connected through the interval.
• No holes: The function should not have undefined points within the interval.

Lacking any of these elements, as in the case here, can invalidate the application of Rolle's Theorem.

Differentiability Explained

Differentiability is about the ability to find the derivative of a function at any point on an interval. For a function to be differentiable at a given point, it must also be continuous there.
In the exercise, the function \(f(x) = x\) is differentiable on the open interval \((0, 1)\), meaning you can calculate its derivative smoothly within this range.
Here are key aspects of differentiability to understand:

• The derivative exists: For each point within the interval, the derivative can be calculated.
• Smooth curves: Differentiability implies no sharp corners or cusps in the function's graph.
• Relation to continuity: If a function is differentiable at a point, it is necessarily continuous at that point.

Although our function does not have a sharp corner or cusp within \((0, 1)\), its derivative \(f'(x)\) is never zero. This aligns with the exercise's finding since a zero derivative isn't mandatory without fulfilling all conditions of Rolle's Theorem.

Understanding Closed Intervals

A closed interval includes its endpoints, written as \([a, b]\), and is essential for certain calculus theorems. Using endpoints in the analysis helps in assessing continuity and other function properties.
In the context of the exercise, Rolle's Theorem specifies conditions over a closed interval \([a, b]\). This means each boundary point, \(a\) and \(b\), is part of the rule set.
For the given function \(f(x)\), to apply Rolle's Theorem, you would need it continuous across \([0, 1]\), but the discontinuity at \(x = 1\) breaks this condition.
Think of a closed interval as:

• Including endpoints: The values at \(a\) and \(b\) are included in the function's evaluation.
• Necessary for certain proofs: Key theorems like Rolle's rely on closed intervals to assert conditions properly.
• Boundary behavior: Understanding behavior at endpoints is vital for determining continuity across the interval.

The failure to maintain continuity on the closed interval \([0, 1]\) is why Rolle's Theorem cannot predict a zero derivative in this scenario.