Problem 6 - Calculus - AP Edition

Question

Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$f(x)=x^{3}+x-4 ;[-1,2]$$

Step-by-Step Solution

Answer

The value of \(c\) that satisfies the Mean Value Theorem is \(c = 1\).

Step 1: Verify Continuity

To apply the Mean Value Theorem, the function must be continuous on the closed interval \([-1, 2]\). The function \(f(x) = x^3 + x - 4\) is a polynomial, which is continuous everywhere, so it is continuous on \([-1, 2]\).

Step 2: Verify Differentiability

The function must also be differentiable on the open interval \((-1, 2)\). Since \(f(x) = x^3 + x - 4\) is a polynomial, it is differentiable everywhere, including on the interval \((-1, 2)\).

Step 3: Confirm MVT Hypotheses

Both conditions of the Mean Value Theorem are satisfied: \(f(x)\) is continuous on \([-1, 2]\) and differentiable on \((-1, 2)\). We can proceed to apply the theorem.

Step 4: Calculate the Function Values at the Endpoints

Evaluate \(f(x)\) at the endpoints of the interval.\[ f(-1) = (-1)^3 + (-1) - 4 = -6 \] \[ f(2) = (2)^3 + 2 - 4 = 6 \]

Step 5: Apply the Mean Value Theorem Formula

According to the Mean Value Theorem, there exists at least one \(c\) in \((-1, 2)\) such that\[ f'(c) = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{6 - (-6)}{2 + 1} = \frac{12}{3} = 4 \]

Step 6: Find the Derivative of the Function

The derivative of the function \(f(x) = x^3 + x - 4\) is\[ f'(x) = 3x^2 + 1 \]

Step 7: Solve for c

We need to find \(c\) such that \(f'(c) = 4\). Hence, solve the equation\[ 3c^2 + 1 = 4 \]\[ 3c^2 = 3 \] \[ c^2 = 1 \] \[ c = \pm 1 \].

Step 8: Validate c in the Interval

Since \(c = -1\) is not in \((-1, 2)\), we discard it.\(c = 1\) is in the interval, so it is the solution.

Key Concepts

Continuity of polynomial functions

Polynomials are among the simplest types of functions you will encounter in calculus. One of the fundamental properties of polynomial functions is their continuity. Continuity, in layman's terms, means that you can draw the graph of the function without lifting your pencil from the paper. For a function to be continuous, it must not have any breaks, jumps, or holes.

The given problem involves a polynomial function:

• Function: \( f(x)=x^{3}+x-4 \)

Since polynomial functions like \( x^3 + x - 4 \) are continuous everywhere on the real number line,

• They are automatically continuous on any interval, including the closed interval \([-1, 2]\).

This ensures the first condition of the Mean Value Theorem is satisfied. Whenever you're verifying continuity for a polynomial, you can confidently state it's continuous across any interval you choose.

Differentiability of polynomial functions

Differentiability is closely related to continuity. If a function is differentiable at a point, it is also continuous at that point. However, the reverse is not always true. Fortunately, for polynomial functions, they are not only continuous everywhere but are also differentiable everywhere.

• Differentiability implies that the function has a derivative at any given point. \( f(x)=x^{3}+x-4 \) is no exception.

In calculus, "differentiability" means you can find the slope of the tangent to the graph at any point. Specifically, this involves calculating the derivative:

• The derivative of \( f(x) = x^3 + x - 4 \) is \( f'(x) = 3x^2 + 1 \).

Within the open interval \((-1, 2)\), the function is differentiable, fulfilling the second condition of the Mean Value Theorem.

Finding critical points

Finding the critical points of a function involves determining where the derivative is either zero or undefined. These points often correspond to where the function's graph has peaks, valleys, or flat points. For our polynomial function, since it is differentiable everywhere, we only need to find where the derivative equals zero or any specified value.

In the context of the Mean Value Theorem, we need to find a specific \(c\) where:

• The derivative \( f'(x) = 3x^2 + 1 \) is equal to a certain value given by the theorem. Here it's 4.

The solution involves solving:

• \( 3c^2 + 1 = 4 \) which simplifies to \( 3c^2 = 3 \).
• Further, \( c^2 = 1 \) gives \( c = \pm 1 \).

Checking the interval \((-1, 2)\),

• Only \( c = 1 \) falls within that range, so it's the point you're looking for that satisfies the Mean Value Theorem.

Note that the ability to derive and solve these equations is key in utilizing mathematical theorems on polynomials.