Problem 11 - Thomas Calculus

Question

a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative. i) \(y=x^{2}-4\) ii) \(y=x^{2}+8 x+15\) iii) \(y=x^{3}-3 x^{2}+4=(x+1)(x-2)^{2}\) iv) \(y=x^{3}-33 x^{2}+216 x=x(x-9)(x-24)\) b. Use Rolle's Theorem to prove that between every two zeros of \(x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) there lies a zero of \(n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\cdots+a_{1}\)

Step-by-Step Solution

Answer

Plot zeros and their derivatives' zeros and observe one zero of derivative between each polynomial zeros, confirming Rolle's Theorem.

Step 1: Calculate Zeros for i)

The polynomial is given by \( y = x^2 - 4 \). To find the zeros, set \( y = 0 \): \( x^2 - 4 = 0 \). This can be factored into \((x-2)(x+2)=0\). The zeros are \( x = 2 \) and \( x = -2 \).

Step 2: Determine First Derivative Zeros for i)

The derivative \( y' = \frac{d}{dx}(x^2 - 4) = 2x \). Setting the derivative to zero gives \( 2x = 0 \), resulting in a zero at \( x = 0 \).

Step 3: Calculate Zeros for ii)

The polynomial is \( y = x^2 + 8x + 15 \). Set \( y = 0 \): \( x^2 + 8x + 15 = 0 \). Factoring gives \((x + 3)(x + 5) = 0\). The zeros are \( x = -3 \) and \( x = -5 \).

Step 4: Determine First Derivative Zeros for ii)

The derivative \( y' = \frac{d}{dx}(x^2 + 8x + 15) = 2x + 8 \). Setting the derivative to zero: \( 2x + 8 = 0 \). Solving gives \( x = -4 \).

Step 5: Calculate Zeros for iii)

Given polynomial \( y = (x+1)(x-2)^2 \). The zeros are directly the roots: \( x = -1 \) and double root at \( x = 2 \).

Step 6: Determine First Derivative Zeros for iii)

Expand \( y = x^3 - 3x^2 + 4 \). The derivative \( y' = 3x^2 - 6x \). Set to zero: \( 3x(x-2) = 0 \). Zeros occur at \( x = 0 \) and \( x = 2 \).

Step 7: Calculate Zeros for iv)

The polynomial is \( y = x(x-9)(x-24) \). Zeros are \( x = 0 \), \( x = 9 \), and \( x = 24 \).

Step 8: Determine First Derivative Zeros for iv)

Expand \( y = x^3 - 33x^2 + 216x \). The derivative is \( y' = 3x^2 - 66x + 216 \). Set to zero: solve \( 3(x^2 - 22x + 72) = 0 \). Factoring gives \((x - 12)(x - 6) = 0\), with zeros at \( x = 6 \) and \( x = 12 \).

Step 9: Use Rolle's Theorem

Rolle's Theorem states if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\) with \( f(a) = f(b) \), then there exists \( c \in (a, b) \) with \( f'(c) = 0 \). Hence, between any two zeros of \( f(x) = x^n + a_{n-1}x^{n-1} + \cdots \) there is at least one zero of its derivative \( f'(x) = n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots \), proving its applicability to all but constant polynomials.

Key Concepts

Polynomial Zeros

A polynomial zero is a value of \( x \) in a polynomial equation \( f(x) = 0 \) where the polynomial evaluates to zero. In simpler terms, these are the values of \( x \) where the graph of the polynomial touches or crosses the x-axis. Finding these zeros involves solving the polynomial equation, often by factoring, using the quadratic formula, or other algebraic methods.

• For the polynomial \( y = x^2 - 4 \), we set it to zero and solve to find \( x = 2 \) and \( x = -2 \), indicating that the graph crosses the x-axis at those points.
• Similarly, for \( y = x^2 + 8x + 15 \), we solve \((x+3)(x+5) = 0\), to find zeros at \( x = -3 \) and \( x = -5 \).
• In higher degree polynomials like \( y = x(x-9)(x-24) \), zeros can be found by identifying the roots from factors, giving \( x = 0, 9, \text{ and } 24 \).

Using these zeros, we can determine the behavior of the polynomial's graph in respect to the x-axis.

Derivative Zeros

Finding zeros of the first derivative \( f'(x) \) of a polynomial gives us information about the critical points where the slope of the tangent to the graph is zero. These are the points where the function might have a local minimum, maximum, or a point of inflection. The process involves differentiating the polynomial and setting the derivative equal to zero to solve for \( x \).

• For a simple polynomial like \( y = x^2 - 4 \), the derivative is \( 2x \). Setting \( 2x = 0 \) gives \( x = 0 \).
• In contrast, the derivative for \( y = x^2 + 8x + 15 \) is \( 2x + 8 \). Setting this to zero yields \( x = -4 \).
• For more complex functions like \( y = x(x-9)(x-24) \), expanding to find the polynomial form helps to derive \( y' = 3x^2 - 66x + 216 \), resulting in zeros at \( x = 6 \) and \( x = 12 \) after solving the quadratic equation.

This analysis is essential as it provides insights into where the function changes direction.

Rolle's Theorem

Rolle's Theorem is a valuable concept in calculus that provides conditions for a polynomial's first derivative to be zero between any two zeros of the polynomial itself. According to Rolle's Theorem, if a polynomial \( f(x) \) is continuous on the interval \([a, b]\), differentiable in \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

• This theorem ensures that between any two consecutive zeros of a polynomial, the derivative must also have a zero.
• For example, in \( y = x^3 - 3x^2 + 4 \), between zeros at \( x = -1 \) and \( x = 2 \), Rolle's Theorem assures there is a zero of the derivative \( y' = 3x^2 - 6x \).
• It supports the idea that the polynomial changes its behavior at some point between these zeros, aligning with the existence of critical points found from the derivative.

Understanding and applying Rolle's Theorem is crucial for analyzing and predicting polynomial functions' graphical behavior between their roots.

Polynomial Functions

Polynomial functions are algebraic expressions made up of variables raised to whole-number powers and coefficients. These functions play a fundamental role in calculus due to their simple structure and widespread application in modeling real-world phenomena.

• A polynomial function's degree indicates the highest power of the variable \( x \), leading to various properties and behaviors.
• For example, quadratic functions (degree 2) like \( y = x^2 - 4 \) consists of terms like \( x^2, 8x, \) and \( 15 \).
• Polynomials can be expressed in factored form such as \((x +1)(x -2)^2\) or expanded form \(y = x^3 - 3x^2 + 4\), and both forms serve different purposes in calculus operations such as differentiation.

Understanding polynomial functions and how to manipulate them is fundamental for solving calculus problems and exploring real-life applications.