Problem 38 - Calculus of a Single Variable

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=2 x^{3}, \quad[0,6]\)

Step-by-Step Solution

Answer

Yes, the Mean Value Theorem can be applied. The value of \(c\) in the open interval \((0, 6)\) that satisfies the equation is \(c = \sqrt{6}\), which lies in the interval.

Step 1: Check Continuity and Differentiability

Since the function \(f(x) = 2x^3\) is a polynomial function, it is continuous and differentiable everywhere, including on the interval \([0, 6]\).

Step 2: Differentiate the Function

Differentiate function \(f(x) = 2x^3\) to get the derivative \(f'(x) = 6x^2\). This derivative function gives the slope of the function at any point \(x\).

Step 3: Apply the Mean Value Theorem

To apply the Mean Value Theorem, set the derivative equal to the average rate of change of the function over the interval \([0, 6]\): \(6x^2 = \frac{f(6) - f(0)}{6 - 0}\). Which simplifies to: \(6x^2 = \frac{2(6)^3 - 2(0)^3}{6}\), thus \(6x^2 = 2(6)^2\), leading to \(x^2 = 6\). Then the solutions of this equation are \(x = \sqrt{6}, -\sqrt{6}\).

Step 4: Select the Suitable value for \(c\) from the Solution

From the Mean Value Theorem, \(c\) should lie in the interval \((0, 6)\). However, in the solution obtained in step 3, only \(\sqrt{6}\) lies in this interval. Hence, we reject \(-\sqrt{6}\).

Key Concepts

Continuity and Differentiability

When it comes to understanding the Mean Value Theorem, grasping the concepts of continuity and differentiability is essential.

A function is said to be continuous at a point if there is no interruption in the graph of the function at that point. This means that the function's value at that point can be directly calculated without any breaks or jumps. For a function to be continuous on a closed interval, it must be continuous at every point within that interval.

Differentiability, on the other hand, is about whether we can calculate a unique tangent at a point on the function's graph. If a function is differentiable at a point, it means that it's smooth and has a well-defined slope at that point. For a function to be differentiable on a closed interval, it must be differentiable at every interior point of that interval and continuous on the entire interval.

The connection between the two concepts becomes clear when we talk about the Mean Value Theorem. It requires that the function in question must be both continuous on the closed interval \[a, b\] and differentiable on the open interval \(a, b\) so that the theorem can be applied. These requirements ensure that there is at least one point in the open interval where the instantaneous rate of change (the derivative) equals the average rate of change over the entire interval. In the given exercise, since we have a polynomial function, which is always continuous and differentiable, we can apply the Mean Value Theorem to \(f(x) = 2x^3\) on the interval \[0, 6\].

Derivative of a Function

In calculus, the derivative of a function is a core concept and is often considered the foundational building block of differential calculus. The derivative represents the rate at which the function's value is changing at any given point. Symbolically, it's denoted as \(f'(x)\) or \frac{dy}{dx}\.To calculate the derivative of a function, you use the principles of differentiation, which can vary based on the type of function involved. For most elementary functions, there are straightforward rules such as the power rule, product rule, and chain rule that can be applied to compute the derivative.

The solution to our exercise involved differentiating a polynomial function using the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Applying this to the function \(f(x) = 2x^3\), we found the derivative to be \(f'(x) = 6x^2\). The derivative \(f'(x)\) not only gives us information on the slope of the tangent line at any point but also is crucial for applying the Mean Value Theorem, as it allowed us to solve for the specific value of \(c\) at which the instantaneous rate of change matches the average rate of change over our interval.

Polynomial Functions

Polynomial functions can be identified by their simplicity and general form of \(f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0\), where the \(a_i\)'s are constants and \(n\) is a non-negative integer. They are composed of one or more terms with variables raised to whole number exponents.

One of the fundamental properties of polynomial functions is their continuity and differentiability at all points on the real number line. This is what makes them relatively easy to work within the context of calculus as compared with functions that can have breaks, cusps, or asymptotes.

In our exercise, \(f(x) = 2x^3\) is an example of a polynomial function. It is the simplicity of polynomial functions that allowed us to perform differentiation effortlessly and apply the Mean Value Theorem directly. As such, knowing the general behavior and properties of polynomials is not only important for understanding their individual behaviors but also for understanding broader concepts in calculus like the Mean Value Theorem.