Problem 149 - Calculus Volume 1

Question

Why do you need differentiability to apply the Mean Value Theorem? Find a counterexample.

Step-by-Step Solution

Answer

Differentiability ensures no cusps; otherwise, MVT fails. Example: \( f(x) = |x| \).

Step 1: Understanding the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). For MVT to hold, the precondition of differentiability ensures that a tangent can be drawn at some point \( c \).

Step 2: Recognizing Importance of Differentiability

Differentiability of a function on an open interval \((a, b)\) implies that the function has no sharp corners or cusps in this interval, ensuring a tangent line can be drawn smoothly. Without differentiability, the function could have discontinuous derivatives, obstructing the condition required by MVT.

Step 3: Identifying a Counterexample

To find a counterexample, you need a function that is continuous but not differentiable on the open interval \((a, b)\). Consider the absolute value function \( f(x) = |x| \) on the interval \([-1, 1]\).

Step 4: Analyzing the Counterexample

The function \( f(x) = |x| \) is continuous everywhere, including on \([-1, 1]\). However, it is not differentiable at \( x = 0 \). If we calculate \( \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - 1}{2} = 0 \), but there is no \( c \) in \((-1, 1)\) where \( f'(c) = 0 \) because the derivative does not exist at \( x = 0 \).

Step 5: Conclusion

Therefore, the Mean Value Theorem cannot be applied where differentiability is not assured. The function \( f(x) = |x| \) acts as a counterexample since it is continuous on \([-1, 1]\) but not differentiable on \((a, b)\), and no such \( c \) exists to satisfy the MVT.

Key Concepts

Differentiability

Differentiability is a key concept in calculus that deals with the ability of a function to have a derivative at each point within its domain. This concept is crucial when applying the Mean Value Theorem (MVT). The MVT requires that a function must be differentiable on an open interval \(a, b\) in order to ensure a tangent can be smoothly drawn.
When a function is differentiable, it means there are no sharp corners, cusps, or vertical tangent lines. A differentiable function promises that at any point within the interval, we can find a derivative, specifically, a subtle slope value.
Without differentiability, the function might have abrupt changes that make the tangent line unrealistic, and this smooth gradient is just what the MVT seeks to find. Differentiability ensures the presence of a point \(c\) in the interval where this gradient is calculated as \frac{f(b) - f(a)}{b - a}\
Guaranteeing lack of differentiability messes up the whole theorem by making it impossible to find such a point.

Continuous Function

A continuous function is one that does not have any breaks, jumps, or holes in its domain. This smooth connection of points is a primary requirement for applying the Mean Value Theorem. Continuous functions mean that for every point \(x\) in the domain, the function approaches the same value from both sides. The MVT states that for a function to be eligible for its application, it must be continuous on a closed interval \([a,b]\).
This continuity, combined with differentiability, is a critical component: while continuity ensures the path is unbroken and smooth, differentiability ensures the slope is well-defined minus abrupt points.
Without the aspect of continuity, no matter how differentiable a function might be over certain parts of its domain, the theorem can't be applied. If a function has a break or jump, imagine stepping over a cliff; you can't have a realistic tangent line, an essential factor needed for the MVT to pinpoint the exact spot it requires.

Counterexample

A counterexample is a specific case that demonstrates the failure of a particular property or theorem. In the context of the Mean Value Theorem, a counterexample is beneficial to show what happens if key conditions are not met.
A classic counterexample for MVT is the absolute value function \( f(x) = |x| \) on the interval \([-1, 1]\).
This function is continuous everywhere, meaning it has no jumps or breaks. However, it is not differentiable at \(x = 0\). At this point, there's a sharp corner, which indicates the absence of a defined tangent.
To calculate using MVT, \frac{f(1) - f(-1)}{1 - (-1)} = 0\, which presumes some \(c\) where the derivative equals zero. However, the derivative of \|f(x) = |x|\ does not exist at \(x = 0\). Thus, no such \(c\) exists in \((-1, 1)\) to satisfy the MVT, proving the necessity of differentiability to apply the theorem correctly.