Problem 13 - Thomas Calculus

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 is not continuous at \( x = 1 \).

Step 1: Understanding the Function

The function \( f(x) \) is a piecewise function. For \( 0 \leq x < 1 \), it is defined as \( f(x) = x \). At \( x = 1 \), it is defined as \( f(x) = 0 \). The function is not continuous at \( x=1 \), as the value changes abruptly from \( f(x) = x \) to \( f(x) = 0 \).

Step 2: Analyzing Rolle's Theorem

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

Step 3: Checking Continuity

For Rolle's Theorem to apply, \( f(x) \) must be continuous on \([0, 1]\). However, \( f(x) \) is not continuous at \( x = 1 \) because the function jumps from \( 1 \) to \( 0 \) at \( x = 1 \).

Step 4: Conclusion from Rolle's Theorem

Since \( f(x) \) is not continuous on \([0, 1]\), one of the conditions for Rolle's Theorem is violated. Thus, the theorem does not apply, and there is no requirement for \( f'(x) \) to be zero anywhere in \((0, 1)\). The derivative, \( f'(x) = 1 \) for \( x \) in \((0, 1)\), is non-zero, which is consistent with the function's definition.

Key Concepts

Piecewise Function

A piecewise function is a type of function defined by distinct expressions over different intervals. In the given problem, the function \( f(x) \) is defined piecewise as follows:

• For \( 0 \leq x < 1 \), \( f(x) = x \)
• At \( x = 1 \), \( f(x) = 0 \)

This means the function behaves differently based on the value of \( x \). Such functions are helpful when modeling scenarios where a "rule" or "behavior" changes at certain points. The abrupt change at \( x = 1 \) in this case indicates that the function has a jump discontinuity at this point.

Continuity

Continuity describes a function without breaks, jumps, or gaps in its domain. A continuous function allows for drawing its graph without lifting your pen from the paper. In our example, \( f(x) = x \) is continuous on \( [0, 1) \). However, at \( x = 1 \), there is a discontinuity because the function value jumps suddenly from 1 to 0, making it not continuous on the closed interval \([0, 1]\). To determine if a piecewise function is continuous at a boundary, check if the left-hand limit, right-hand limit, and the function value all equal the same. Here, they do not, causing the discontinuity. This mismatch is the crux of why Rolle’s Theorem fails to apply.

Differentiable

A function is differentiable at a point if it has a defined derivative at that point. For most continuous functions, differentiability implies smoothness without any sharp corners. In the case of \( f(x) = x \) for \( 0 \leq x < 1 \), the function is differentiable, and its derivative \( f'(x) = 1 \) remains constant and non-zero. Differentiability means we can compute the slope of the tangent line at any point within that part of the function. Even though the function is not continuous at \( x = 1 \), it remains differentiable over the interval \((0, 1)\). This highlights an important aspect: a function can be differentiable without fulfilling all continuity conditions everywhere in its domain.

Function Analysis

Function analysis involves examining a function’s properties and behavior such as continuity, differentiability, and potential application of calculus theorems. A critical part of this analysis is verifying where these properties hold or fail.For the problem at hand, Rolle's Theorem can’t be applied because:

• The function must be continuous on \([a, b]\); yet, \( f(x) \) is not continuous at \( x = 1 \).
• The function must be differentiable on \((a, b)\); it is, however, differentiable in \((0, 1)\).
• \( f(a) = f(b) \); while \( f(0) = f(1) = 0 \), the failure in continuity checks stops Rolle’s from applying.

This analysis confirms that between 0 and 1, despite the derivative being constant (\( f'(x) = 1 \)), the result agrees with \( f(x) \) not needing a point where \( f'(x) = 0 \), due to the violated continuity condition.