Mean Value Theorem

Summary

This video explains the Mean Value Theorem (MVT) in calculus, detailing its conditions and practical applications. It highlights that for a continuous and differentiable function over a closed interval, there exists a point where the instantaneous rate of change (slope of the tangent line) equals the average rate of change (slope of the secant line). The video provides several examples, including polynomial functions, absolute value functions, and square root functions, demonstrating how to find the value of 'c' that satisfies the theorem. Key concepts include differentiability, continuity, and the relationship between secant and tangent slopes.

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Q & A

What are the conditions for applying the Mean Value Theorem (MVT)?

The function must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b).

What does the Mean Value Theorem state about the relationship between the instantaneous rate of change and the average rate of change?

The MVT states that there exists a point c in the open interval (a, b) where the instantaneous rate of change (the derivative at c) is equal to the average rate of change over the interval [a, b].

What is the difference between a secant line and a tangent line in the context of the MVT?

The secant line connects two points on the curve and represents the average rate of change between those points. The tangent line touches the curve at a single point and represents the instantaneous rate of change at that point.

In the provided example with the function f(x) = x^2 - 4x + 1 on the interval [1, 5], what is the value of c that satisfies the Mean Value Theorem?

The value of c that satisfies the MVT is c = 3, as it lies between 1 and 5 and ensures the slope of the tangent line equals the slope of the secant line.

Why can't the Mean Value Theorem be applied to the function f(x) = x^(2/3) on the interval [0, 1]?

The function is not differentiable at x = 0 because it has a cusp there, which means it cannot satisfy the differentiability condition for the Mean Value Theorem on the open interval (0, 1).

Why does the absolute value function f(x) = |4x - 5| on the interval [0, 2] fail to meet the criteria for applying the MVT?

The absolute value function has a sharp corner at x = 5/4, which makes the function non-differentiable at that point. Since this point lies inside the interval [0, 2], the MVT cannot be applied.

What is the first step in applying the Mean Value Theorem to a function?

The first step is to verify that the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If both conditions are met, you can apply the MVT.

How do you calculate the value of c in the equation f'(c) = (f(b) - f(a)) / (b - a)?

To calculate the value of c, first find the derivative of the function, then calculate the average rate of change (f(b) - f(a)) / (b - a). Afterward, solve the equation f'(c) = (f(b) - f(a)) / (b - a) for c.

What happens if a function is continuous but not differentiable at some point in the interval?

If the function is continuous but not differentiable at some point in the interval, the Mean Value Theorem cannot be applied, as differentiability is a necessary condition on the open interval (a, b).

In the problem with f(x) = sqrt(x - 4) on the interval [4, 8], what is the value of c that satisfies the MVT?

The value of c that satisfies the MVT is c = 5, as it lies between 4 and 8, and ensures the slope of the tangent line equals the slope of the secant line.