Problem 66 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The polynomial has one local minimum at \( x = 0 \) and no local maxima.

Step 1: Understanding the Polynomial

We are given the polynomial \( y = (x^2 - 2)^3 \). Our goal is to graph it and determine the number of local maxima and minima. This is a degree 6 polynomial (since when expanded, the highest power of \( x \) will be 6). It is likely to have multiple turning points, which include local maxima and minima.

Step 2: Finding the Derivative

To find the local maxima and minima, we need to identify the critical points by taking the derivative and setting it to zero. The derivative \( y' \) is found using the chain rule: \[ y' = 3(x^2 - 2)^2 \cdot (2x) = 6x(x^2 - 2)^2 \].

Step 3: Solving Critical Points

Set the derivative equal to zero: \[ 6x(x^2 - 2)^2 = 0 \].This equation is satisfied when \( x = 0 \) or \( x^2 - 2 = 0 \). Solving \( x^2 - 2 = 0 \), we get \( x = \pm\sqrt{2} \). Thus, the critical points are \( x = 0, \sqrt{2}, \) and \( -\sqrt{2} \).

Step 4: Determining Nature of Critical Points

We use the second derivative test to determine if these critical points are maxima or minima. First, calculate the second derivative:\[ y'' = \frac{d}{dx}(6x(x^2 - 2)^2) \].Using the product and chain rule, \[ y'' = 6(x^2 - 2)^2 + 24x^2(x^2 - 2) \].

Step 5: Evaluating Second Derivative at Critical Points

Evaluate \( y'' \) at each critical point:- At \( x = 0 \): \( y'' = 6(-2)^2 = 24 \) (\( y'' > 0 \), local minimum).- At \( x = \sqrt{2} \) and \( x = -\sqrt{2} \): - \( y'' = 0 \) at these points, the second derivative test is inconclusive.To resolve, consider the first derivative test or analyze the graph.

Step 6: Graphing and Visual Inspection

Graph \( y = (x^2 - 2)^3 \) to confirm the analysis. The function exhibits symmetry about the y-axis. You can plot points or use graphing software. Visually, at \( x=\sqrt{2} \) and \( x=-\sqrt{2} \), the curve appears to have points of inflection, not local maxima/minima. At \( x = 0 \), there is a local minimum.

Key Concepts

Local Maxima

In the context of polynomial graphing, a local maximum refers to a point where the function peaks locally. In other words, within a small region, it is the highest point when compared to nearby values. If you're climbing a hill, the local maximum would be the top of that particular hill, even if there might be taller hills elsewhere on the graph.

To find these local maxima mathematically, we typically use the first and second derivative tests.

• The first derivative test involves finding the critical points, where the derivative equals zero or does not exist. At these points, the function may change from increasing to decreasing, indicating a peak or local maximum.
• The second derivative test helps confirm whether a critical point is actually a maximum. If the second derivative at that point is less than zero, it signals that the point is a local maximum.

In the specific exercise of the polynomial \( y = (x^2 - 2)^3 \), analysis determined there are no local maxima. These points are typically identified through careful inspection of the graph and verifying through derivative tests.

Local Minima

A local minimum is, quite simply, the lowest point in a particular section of a graph, much like the bottom of a valley between hills. In a neighborhood of points, it represents the least value which is larger than the surrounding points.

Using the derivative approach, you can find potential local minima by identifying the critical points.

• If the first derivative equals zero, it's a signal that this point may be either a hilltop or a valley bottom.
• The second derivative again can clarify: if the second derivative value at a critical point is more than zero, then you have yourself a local minimum there.

For our polynomial in the original problem, certainly, the point \( x = 0 \) is where a local minimum occurs because the second derivative is positive, confirming it's a valley not a peak.

By graphing the polynomial \( y = (x^2 - 2)^3 \), we see that at this point, the curve dips to its lowest in the local neighborhood.

Critical Points

Critical points are fundamental in determining the shape and features of a polynomial graph as they suggest where the slope of the curve is zero or undefined. In simpler terms, these are the places where a graph "stops" going up or down—creating peaks, valleys, or flat spots.

Identifying critical points is key in finding the local maxima, minima, or points of inflection. Here's how:

• To locate critical points, first calculate the derivative of the function, set it to zero, and solve for the variable.
• These solutions are potential critical points, but further analysis will determine their precise nature.

In the polynomial \( y = (x^2 - 2)^3 \), the derivative \( y' = 6x(x^2 - 2)^2 \) was used to find the critical points at \( x = 0, \sqrt{2}, \) and \( -\sqrt{2} \). Each point requires further testing to confirm whether it's a maxima, minima, or points of inflection.

Knowing and testing these critical points are crucial steps in graphing polynomials efficiently, ensuring all local characteristics are accurately represented.