Problem 27 - Calculus Single Variable

Question

In each of Exercises \(23-28\), verify that the hypotheses of the Rolle's Theorem hold for the given function \(f\) and interval I. The theorem asserts that \(f^{\prime}(c)=0\) for some \(c\) in \(I .\) Find such a \(c\). $$ f(x)=x^{3}+4 x^{2}+x-6, \quad I=[-2,1] $$

Step-by-Step Solution

Answer

The point \(c\) where \(f'(c) = 0\) is approximately \(c = \frac{-4 + \sqrt{13}}{3}\).

Step 1: Check Continuity

For Rolle's Theorem, the function must be continuous on the closed interval \([-2, 1]\). The function \(f(x) = x^3 + 4x^2 + x - 6\) is a polynomial, and polynomials are continuous everywhere. Thus, \(f(x)\) is continuous on \([-2, 1]\).

Step 2: Check Differentiability

The function must be differentiable on the open interval \((-2, 1)\). Since \(f(x)\) is a polynomial, it is differentiable everywhere. Therefore, \(f(x)\) is differentiable on \((-2, 1)\).

Step 3: Verify f(a) = f(b)

For Rolle's Theorem to apply, \(f(-2)\) must equal \(f(1)\). Evaluate \(f(-2)\):\[ f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6 = -8 + 16 - 2 - 6 = 0 \]Now, evaluate \(f(1)\):\[ f(1) = 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0 \]Since \(f(-2) = f(1) = 0\), the condition \(f(a) = f(b)\) holds.

Step 4: Find c such that f'(c) = 0

Compute the derivative of \(f(x)\):\[ f'(x) = 3x^2 + 8x + 1 \]Set the derivative equal to zero to find critical points:\[ 3x^2 + 8x + 1 = 0 \]Solve for \(x\) using the quadratic formula, where \(a = 3\), \(b = 8\), and \(c = 1\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{-8 \pm \sqrt{64 - 12}}{6} = \frac{-8 \pm \sqrt{52}}{6} = \frac{-8 \pm 2\sqrt{13}}{6} \]Simplify:\[ x = \frac{-4 \pm \sqrt{13}}{3} \]These are the critical points. We check which of them lies in the interval \([-2, 1]\). The root \(x = \frac{-4 + \sqrt{13}}{3}\) is approximately 0.53, which lies in \([-2, 1]\).

Key Concepts

Continuity

A function must be continuous over an interval for Rolle's Theorem to apply. Continuity means that you can draw the function without lifting your pen off the paper. This concept is crucial because it ensures there are no breaks, jumps, or holes in the graph of the function.
For the given function, which is a polynomial, continuity is guaranteed on any interval. Polynomials, like our function \( f(x) = x^3 + 4x^2 + x - 6 \), are smooth and unbroken graphs not only on \([-2, 1]\) but on all real numbers. This means it passes the first hurdle of Rolle's Theorem easily.

• Continuous functions are defined at every point in the interval.
• No sudden jumps or abrupt changes.
• Essential for the application of many calculus theorems.

Differentiability

For Rolle's Theorem to be applicable, a function must also be differentiable over an open interval. Differentiability tells us that there is a tangent line at every point on the curve and that the function has a derivative at each point.
For our exercise, differentiability was straightforward to check since polynomials are differentiable everywhere, much like their continuous nature. That means \( f(x) = x^3 + 4x^2 + x - 6 \) is differentiable over \((-2, 1)\) without exception.

• Differentiable functions have a defined slope or rate of change.
• No sharp corners or cusps in differentiable functions.
• Smooth and predictable behavior.

Polynomial Function

Polynomials are a type of mathematical function that is composed of terms of the form \( ax^n \), where \( a \) is a coefficient and \( n \) is a non-negative integer. They are one of the simplest types of functions in calculus due to their continuous and differentiable properties.
In this exercise, we worked with the polynomial \( f(x) = x^3 + 4x^2 + x - 6 \). Here’s why polynomials are straightforward yet powerful:

• They are continuous on any closed interval.
• They are differentiable on any open interval.
• Their derivatives are easily computed, as seen when deriving \( f'(x) = 3x^2 + 8x + 1 \).
• Polynomials can be of any degree, with linear, quadratic, and cubic terms as common examples.

Critical Points

Critical points occur where a function's derivative is zero or undefined, marking potential local maxima, minima, or points of inflection. They are crucial in calculus because they help identify behavior changes in the graph of a function.
In the context of Rolle's Theorem, we found the critical points by setting the derivative \( f'(x) = 3x^2 + 8x + 1 \) equal to zero. Solving the quadratic equation gave us the critical points:
\[ x = \frac{-4 \pm \sqrt{13}}{3} \]
Evaluating these gave values like 0.53, which lies within the interval \([-2, 1]\). Therefore, this is where the behavior of the function reaches a peak or a valley on that interval.

• Local maxima and minima are where the graph peaks or valleys.
• Points of inflection mark changes in concavity.
• Identifying critical points is essential for graph analysis and optimization problems.