Problem 61 - Algebra and Trigonometry

Question

\(59-68\) Graph the polynomial and determine how many local maxima and minima it has. $$ y=x^{3}-x^{2}-x $$

Step-by-Step Solution

Answer

One local maximum and one local minimum.

Step 1: Understanding the Polynomial

The polynomial given is \( y = x^3 - x^2 - x \). It is a cubic polynomial, which typically means it may have up to two turning points where local maxima and minima can occur.

Step 2: Find the Derivative

First, calculate the first derivative of the polynomial, as it will help in finding the critical points where maxima or minima might occur. \[ y' = \frac{d}{dx}(x^3 - x^2 - x) = 3x^2 - 2x - 1 \]

Step 3: Find Critical Points

Set the derivative equal to zero to find the critical points.\[ 3x^2 - 2x - 1 = 0 \]Solving this quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -2 \), \( c = -1 \), we use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} \]\[ x = \frac{2 \pm 4}{6} \]Solving for \( x \), gives us \( x = 1 \) and \( x = -\frac{1}{3} \). These are the critical points.

Step 4: Determine the Nature of Critical Points

To determine if these critical points are maxima or minima, check the second derivative.\[ y'' = \frac{d}{dx}(3x^2 - 2x - 1) = 6x - 2 \]Evaluate the second derivative at each critical point:For \( x = 1 \):\[ y''(1) = 6 \times 1 - 2 = 4 \]Since \( y''(1) > 0 \), this is a local minimum.For \( x = -\frac{1}{3} \):\[ y''(-\frac{1}{3}) = 6 \times (-\frac{1}{3}) - 2 = -2 - 2 = -4 \]Since \( y''(-\frac{1}{3}) < 0 \), this is a local maximum.

Step 5: Graph the Polynomial

Plot the graph of the polynomial \( y = x^3 - x^2 - x \) to visually confirm the locations of the local max and min. The graph will show a high point at \( x = -\frac{1}{3} \) (local maximum) and a low point at \( x = 1 \) (local minimum).

Key Concepts

Cubic Polynomials

Cubic polynomials are mathematical expressions of degree three, meaning the highest power of the variable is cubed. The general form is \( ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are constants, and \( a eq 0 \). These polynomials can have up to three roots and display interesting behaviors such as curves and inflections.

• Roots: These can be found by solving the equation \( ax^3 + bx^2 + cx + d = 0 \). Depending on the discriminant, a cubic polynomial can have one real root and two complex ones, or three real roots.
• Graphical Behavior: The graph of a cubic polynomial can have distinct inflection points where the direction of curvature changes.
• Turning Points: Uniquely, cubic polynomials can feature up to two turning points, known as local maxima and minima, but not every cubic has both types. These are crucial for understanding the behavior of the graph.

Local Maxima and Minima

Local maxima and minima are key concepts in calculus, representing the highest and lowest points on a function's graph in a given interval. For cubic polynomials, these points occur where the derivative is zero.

• Local Maximum: This is a point where the function reaches the greatest value in a neighboring region. It's where the slope changes from positive to negative.
• Local Minimum: Here, the function attains its least value locally, with the slope changing from negative to positive.

To determine if a point is a local maximum or minimum, we use the first and second derivatives. Specifically, if the second derivative at a critical point is positive, it's a local minimum; if it's negative, it's a local maximum. In the given exercise, the polynomial \( y = x^3 - x^2 - x \) has critical points showing both a local max and min.

Derivative Calculations

Derivative calculations are essential for identifying critical points and analyzing a function's behavior. The derivative of a function gives us the rate at which the function's value changes.

• Finding the Derivative: For the polynomial \( y = x^3 - x^2 - x \), the derivative is \( y' = 3x^2 - 2x - 1 \). This derivative helps locate critical points, where the slope is zero.
• Critical Points: These occur where the derivative equals zero. Solving \( 3x^2 - 2x - 1 = 0 \) allows us to find where the function's rate of change is null.
• Second Derivative: Calculating \( y'' = 6x - 2 \) is crucial for determining the nature of the critical points found. It helps ascertain whether the points represent maxima or minima.

Quadratic Formula

The quadratic formula is a fundamental tool used to find the solutions of a quadratic equation \( ax^2 + bx + c = 0 \). This formula is essential when working with derivatives that are quadratic forms.
The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• Application: In the exercise, the derivative \( 3x^2 - 2x - 1 \) is set to zero to find critical points. Importantly, this shows the application of the quadratic formula.
• Procedure: Plug the coefficients \( a = 3 \), \( b = -2 \), \( c = -1 \) into the formula to solve for \( x \).
• Result: It yields \( x = 1 \) and \( x = -\frac{1}{3} \), indicating the points where the polynomial changes direction.

The quadratic formula is indispensable for finding solutions quickly and is integral in calculus and algebra for solving complex expressions.