Problem 13 - Thomas Calculus in SI Units

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

The discontinuity at \( x=1 \) invalidates the conditions for Rolle's Theorem.

Step 1: Understand Rolle's Theorem

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

Step 2: Identify the Function Properties

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

Step 3: Verify Continuity and Differentiability Conditions

The function \( f(x) \) is continuous on \([0, 1)\) since \( f(x) \) is simply \( x \) from \( [0, 1) \), which is a continuous expression. However, it is not defined as continuous at \( x=1 \), disrupting continuity on the closed interval \([0, 1]\) required for Rolle's Theorem to apply.

Step 4: Assess if Rolle’s Theorem Applies

Rolle's Theorem requires the function to be continuous on the closed interval, \([0, 1]\), and the function value at the endpoints to be the same. Since the function is not defined at \( x = 1 \), the continuity requirement on \([0, 1]\) is not met. Hence, Rolle's Theorem does not apply here.

Key Concepts

Continuous Function

Understanding the concept of a continuous function is vital in distinguishing whether Rolle's Theorem applies to a given scenario. A function is said to be continuous on a closed interval \( [a, b] \) if there are no breaks, jumps, or holes in its graph within that interval.
This means that for every point in the interval, the function approaches the value from either side without any interruption.For the function given in the exercise, it is represented as \( f(x) = x \) on the interval \( [0, 1) \), which means it is continuous within this part of the interval because a polynomial function like \( x \) does not have any breaks or holes.
However, at the point \( x = 1 \), \( f(x) \) takes a different value; it becomes 0, and thus, creates a jump in the graph.

• The interval \( [0, 1) \) is where the function is continuous.
• At \( x = 1 \), the value does not match the continuity definition, causing discontinuity over the closed interval.

Hence, although parts of the function are continuous, it fails to be continuous over the entire closed interval \( [0, 1] \), which is essential for Rolle's Theorem.

Differentiable Function

A differentiable function is one that can be differentiated, meaning it has a derivative at every point in the open interval \( (a, b) \).
Differentiability implies that the function is "smooth" and has no sharp corners or cusps in the interval.In our function, \( f(x) = x \) within the interval \( (0, 1) \) is a simple linear function.
Linear functions are differentiable everywhere in their domain as their derivative exists at all points, defined as the constant slope of the line.

• For the interval \( (0, 1) \), the polynomial part \( f(x) = x \) is differentiable with a derivative equal to 1 throughout.
• No corners or discontinuities exist within \( (0, 1) \), hence the function is smooth in this region.

Since the differentiability condition is met for \( (0, 1) \), the issue isn't with the function's differentiability but rather its continuity at the endpoints on the closed interval.

Piecewise Function

Piecewise functions are defined by different expressions based on the input ranges they are applied to.
In this case, the function is defined differently for segments of its domain, making it a piecewise function.The given function illustrates this feature:

• For \( 0 \leq x < 1 \), \( f(x) = x \). This is a linear function part.
• At \( x = 1 \), \( f(x) = 0 \), a constant function.

The idea of a piecewise function can create challenges in applying theorems like Rolle's, which require certain conditions across the full domain.
Although each piece is continuous within its domain and differentiable on its open intervals, the switch in definition at the endpoint affects the overall application of continuity needed for the closed interval analysis in Rolle's Theorem. Therefore, even though each individual piece exhibits necessary mathematical properties, the combined function fails to meet the prerequisites necessary to utilize Rolle's Theorem over the full closed interval \( [0, 1] \).