Let ƒ(x) = x²⸍³ . Show that there is no value of c in the interval (-1, 8) for which ​​ƒ' (c) = (ƒ(8) - ƒ (-1)) / (8 - (-1)) and explain why this does not violate the Mean Value Theorem.

Verified step by step guidance

1. First, let's understand the Mean Value Theorem (MVT). The MVT states that if a function ƒ is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that ƒ'(c) = (ƒ(b) - ƒ(a)) / (b - a).
2. Calculate ƒ(8) and ƒ(-1) using the given function ƒ(x) = x²⸍³. Substitute x = 8 and x = -1 into the function to find these values.
3. Next, compute the average rate of change of the function over the interval [-1, 8] using the formula (ƒ(8) - ƒ(-1)) / (8 - (-1)).
4. Find the derivative of the function ƒ(x) = x²⸍³. Use the power rule for differentiation, which states that if ƒ(x) = xⁿ, then ƒ'(x) = n*xⁿ⁻¹.
5. Check the differentiability of ƒ(x) = x²⸍³ on the interval (-1, 8). Since the function is not differentiable at x = 0 (due to the fractional exponent), the conditions of the Mean Value Theorem are not fully satisfied, explaining why there is no value of c in (-1, 8) for which ƒ'(c) equals the average rate of change.

Key Concepts

Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that the derivative at that point equals the average rate of change of the function over the interval. This theorem is fundamental in understanding the relationship between a function's behavior and its derivative.

Differentiability and Continuity

A function is differentiable at a point if it has a defined derivative there, which implies that the function is also continuous at that point. However, a function can be continuous without being differentiable. In the context of the MVT, if a function is not differentiable at any point in the interval, it cannot satisfy the conditions of the theorem, which is crucial for determining the existence of such a point c.

Behavior of the Function ƒ(x) = x²/3

The function ƒ(x) = x²/3 is continuous everywhere but is not differentiable at x = 0, where it has a cusp. This means that while the function meets the continuity requirement of the MVT on the interval [-1, 8], it fails the differentiability requirement at c = 0. Therefore, there is no value of c in (-1, 8) that satisfies the MVT, which does not violate the theorem since the conditions are not fully met.