Problem 9 - 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 de- rivative 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 \(f(x)\) is not continuous at \(x=1\).

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 exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\). This implies that the graph of \(f(x)\) has a horizontal tangent somewhere within that interval.

Step 2: Analyzing the Function

The function \(f(x)\) is defined as \(f(x) = x\) for \(0 \leq x < 1\) and \(f(x) = 0\) for \(x = 1\). This function is continuous over \([0, 1]\), but at \(x = 1\), it has a point discontinuity because it jumps from \(x\) to \(0\). Therefore, \(f(x)\) is not continuous over the entire interval \([0, 1]\).

Step 3: Applying Rolle's Theorem Conditions

Rolle's Theorem requires the function to be continuous over the closed interval; however, \(f(x)\) is not continuous at \(x = 1\). The jump from \(f(1 - \epsilon) = 1 - \epsilon\) to \(f(1) = 0\) violates the continuity requirement. Therefore, we cannot apply Rolle's Theorem.

Step 4: Examining the Derivative

For \(0 < x < 1\), the derivative of \(f(x) = x\) is \(f'(x) = 1\). Since the derivative is constant and equal to 1 on \((0, 1)\), it is never zero within this interval. The derivative matches what we expect given the linear nature of \(f(x) = x\) between \((0, 1)\).

Step 5: Conclusion

Rolle's Theorem cannot be applied here because the function \(f(x)\) is not continuous at \(x = 1\) within the closed interval \([0, 1]\). Without continuity across the entire interval, the conditions of the theorem are not satisfied, and thus there is no guarantee, nor expectation, for the derivative to be zero on \((0, 1)\).

Key Concepts

Derivative

The derivative of a function represents the rate at which the function's value changes as the input changes. It is essentially the slope of the tangent line to the graph of the function at any given point.
In the context of the provided function, for the interval \((0, 1)\), we are focusing on the derivative of \(f(x) = x\).
When we differentiate \(f(x) = x\), we find that \(f'(x) = 1\) for \(0 < x < 1\).

• This indicates that the slope of the function is a constant 1 between \(0\) and \(1\).
• Since the derivative equals 1, it never reaches zero between \(0\) and \(1\).

Therefore, the absence of any point where the derivative is zero within this interval explains why the function does not align with certain expectations from Rolle's Theorem under normal conditions.

Continuity

Continuity refers to a function being unbroken and smoothly connected over an interval. For a function to be considered continuous at a point, it must not have any breaks, jumps, or abrupt changes.
In the problem at hand, the function \(f(x)\) demonstrates a notable discontinuity at \(x = 1\).

• For \(0 \leq x < 1\), \(f(x) = x\) is continuous since it progresses smoothly without interruption.
• However, at \(x = 1\), \(f(x)\) suddenly changes from \(1\) to \(0\), creating a jump, which forms a point of discontinuity.

This break at \(x = 1\) means the function doesn't meet the required continuity conditions of the closed interval \([0, 1]\) for applying Rolle's Theorem, as it's not continuous everywhere within \([0, 1]\). This is why we can't use the theorem here, despite seeing \(f(0) = f(1) = 0\).

Differentiability

Differentiability indicates whether a function has a derivative at each point in its interval. For a function to be differentiable at a point, it must also be continuous there, but the reverse isn't always true.
For the function \(f(x)\) defined as above, differentiability applies within the interval between, but not including, \(0\) and \(1\).

• On \((0,1)\), \(f(x) = x\) is differentiable because it has a constant slope of \(1\).
• However, at the endpoint \(x = 1\), differentiability is not defined due to the discontinuity discussed earlier.

Thus, even though \(f(x)\) is differentiable on an open interval \((0, 1)\), the breach in continuity at \(x = 1\) restricts using Rolle's Theorem, which demands differentiability across \((a, b)\) and continuity from \([a, b]\). This highlights how the concepts of continuity and differentiability intersect and affect the applicability of calculus theorems.