Problem 3 - Calculus of a Single Variable

Question

Explain why Rolle's Theorem does not apply to the function even though there exist \(a\) and \(b\) such that \(f(a)=f(b) .\) \(f(x)=1-|x-1|, \quad[0,2]\)

Step-by-Step Solution

Answer

Rolle's Theorem doesn't apply to the function \(f(x)=1-|x-1|\) on the interval [0,2] because, despite the function being continuous on [0,2] and satisfying \(f(0)=f(2)\), it is not differentiable at \(x=1\) which is in the open interval (0,2). Rolle's Theorem requires the function to be differentiable on the entire open interval (a,b), and thus this condition is not fulfilled.

Step 1: Recall the conditions for Rolle's Theorem

Rolle's theorem, a special case of the Mean Value Theorem, applies for a function that is continuous on the interval [a, b], differentiable on the open interval (a, b), and for which \(f(a)=f(b)\).

Step 2: Analyze the given function

The function \(f(x)=1-|x-1|\) is continuous on the interval [0,2] and we can verify that \(f(0)=f(2)\) by substituting \(x=0\) and \(x=2\) into the function: \(f(0)=1-|0-1|=1-|-1|=1-1=0\) and \(f(2)=1-|2-1|=1-|-1|=1-1=0\). So, \(f(a)=f(b)\) condition is met.

Step 3: Examine the differentiability of the given function

While the function is continuous on [0,2] and satisfies \(f(a)=f(b)\), it is not differentiable on the open interval (0,2). The absolute value function |-x+1| is non-differentiable at the point where x=1 (the vertex of the V-shape) because it has a corner there. So, the function \(f(x)\) fails to meet this essential condition of Rolle's Theorem.

Step 4: Concluding Remarks

Thus, by observing that the function \(f(x)=1-|x-1|\) is not differentiable over the full interval (0,2), that is, it fails at \(x=1\), which falls within said interval, it is clear why Rolle's Theorem doesn't apply to it, even though the other conditions are satisfied.

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental principle in calculus that helps us understand how functions behave over an interval. It states that if a function is continuous on a closed interval \[a, b\] and differentiable on the open interval \(a, b\), there is at least one point \(c\) in that interval where the instantaneous rate of change (derivative) equals the average rate of change over \[a, b\]. In simple terms, the function's slope at some point is exactly the same as the slope of the line connecting \(f(a)\) and \(f(b)\).
For Rolle's Theorem, a special case of MVT, the additional condition \(f(a) = f(b)\) must be satisfied. This means there should be at least one point in between where the derivative is zero, indicating a horizontal tangent line.

differentiability

Differentiability is an essential concept in calculus that describes whether a function has a derivative at each point within an interval. A function is differentiable at a point if it is smooth and has a tangent line without any breaks, sharp corners, or cusps at that point. For a function to be differentiable over an entire interval, it must be smooth everywhere in that interval.
If a function involves an absolute value, it often introduces a point where the function may not be differentiable. For example, the function \(f(x) = 1 - |x-1|\) has a corner at \(x=1\), making it non-differentiable at that point, despite being differentiable elsewhere on the interval. This lack of differentiability at even one point prevents the application of the Mean Value Theorem or Rolle's Theorem over the entire interval.

continuity

Continuity ensures that a function doesn't have any breaks, jumps, or gaps, and can be drawn without lifting the pen from the paper. A function is continuous on a closed interval \[a, b\] if you can pass through every point within that interval smoothly. For the Mean Value Theorem and Rolle's Theorem to apply, the function must be continuous over the interval including its endpoints.
The function \(f(x) = 1 - |x-1|\) is continuous on the interval \[0, 2\], meaning there are no disruption points in this range. Continuity alone isn't enough for Rolle's Theorem, as differentiability must also be present, which we saw isn't fully met in this case.

absolute value function

The absolute value function, familiar to many, is denoted as \(|x|\) and is used to measure the distance of a number from zero on the number line. It produces a 'V' shape graph, known for having a sharp corner at the point where the function value becomes zero. This characteristic can lead to problems with differentiability.
For the function \(f(x) = 1 - |x-1|\), the absolute value component \(|x-1|\) causes it to have a corner at \(x = 1\). This corner is why the function isn't differentiable at that point, despite being well-behaved elsewhere. Understanding the behavior of the absolute value function helps explain why Rolle's Theorem cannot be applied in such cases, due to a failure in meeting the differentiability requirement over the entire interval.