Problem 32 - Essential Calculus

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 \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). $$ f(x)=x\left(x^{2}-x-2\right),[-1,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem can be applied to the function and the value of \(c\) in the open interval \((-1, 1)\) that satisfies the Mean Value Theorem is \(-\frac{1}{6}\).

Step 1: Verify if the function meets MVT conditions

The function \(f(x) = x(x^2 - x - 2)\) is a polynomial function, and polynomial functions are continuous and differentiable for all real numbers. Thus, it is continuous on \([-1, 1]\) and differentiable on \((-1, 1)\). So, the Mean Value Theorem applies here.

Step 2: Compute Function's Derivative

Take derivative of the function \(f(x)\) using product and chain rules. The derivative \(f'(x)\) of the function \(f(x)\) can be found as follows: \(f'(x) = 3x^2 - 2x - 2\).

Step 3: Apply the Mean Value Theorem

Now, according to the Mean Value Theorem, there should exist value(s) of \(c\) in the interval \((-1, 1)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Compute \(f(b) - f(a)\) and substitute \(b = 1\) and \(a = -1\) into the function \(f(x) = x(x^2 - x - 2)\). Hence, \(f(1) - f(-1) = 1 - 4 = -3. So, f'(c) = \frac{f(1)-f(-1)}{1 - (-1)} = \frac{-3}{2}\).

Step 4: Solve for c

To find c, equate \(f'(c)\) to \(\frac{-3}{2}\) and solve for \(c\): \(3c^2 - 2c - 2 = \frac{-3}{2}\). This simplifies to: \(3c^2 - 2c - \frac{1}{2} = 0\). We can solve this quadratic equation to find the values of \(c\). The solutions turn out to be \(c = 1, c = -\frac{1}{6}\). Only -1/6 lies in the open interval (-1,1).

Key Concepts

Continuity in Calculus

Understanding continuity in calculus is crucial for applying various theorems, including the Mean Value Theorem (MVT). A function is said to be continuous on an interval if it has no breaks, jumps, or holes in that interval. In simpler terms, you should be able to draw the graph of the function without lifting your pencil within that interval.

For the MVT to be applicable to a function on a closed interval \[a, b\], the function must be continuous on that interval. Polynomial functions, like the one given in our exercise \(f(x) = x(x^2 - x - 2)\), are inherently continuous everywhere. That's because they are made up of terms that are simply constants or powers of the variable, making their graphs smooth and unbroken lines or curves.

Differentiability

Differentiability is closely linked to continuity but includes an extra condition; a function must not only be continuous but also have a defined derivative at every point in an interval to be considered differentiable there. A differentiable function allows us to calculate the rate at which it's changing at any point within an interval, which is the essence of calculus.

In the context of the MVT, the function in question must be differentiable on the open interval \(a, b\). Since the derivative of a polynomial function exists at all points, our given function \(f(x) = x(x^2 - x - 2)\) meets this requirement as well, allowing us to apply the MVT and find the specific value or values of \(c\) in the interval that satisfy the theorem's equation.

Polynomial Functions

Polynomial functions are algebraic expressions that involve only non-negative integer powers of the variable and have a variety of important properties. As mentioned, they are continuous and differentiable everywhere, making them very predictable and their behavior easy to analyze mathematically.

The function from our exercise, \(f(x) = x(x^2 - x - 2)\), is a cubic polynomial. Polynomials have nice, smooth graphs, which is why they don't exhibit any sudden jumps or vertical asymptotes. These attributes ensure that the conclusions drawn from applying the Mean Value Theorem to polynomials are reliable and meaningful.

Derivatives

A derivative represents the rate of change of a function with respect to its independent variable. This is analogous to calculating the slope of the tangent line at any point on the function's graph. Derivatives are fundamental to calculus because they give us insights into the function's behavior: whether it is increasing, decreasing, or staying constant at a particular point.

In the case of our exercise, calculating the derivative of the polynomial function \(f(x) = x(x^2 - x - 2)\) gives us a new function \(f'(x)\) that tells us the slope at any point \(x\). It's this derivative that we set equal to the average rate of change over the interval \[a, b\] to find the \(c\) values in the Mean Value Theorem. The resulting equation can then be solved to determine where the instantaneous rate of change equals the average rate of change on the interval.