Problem 52 - Precalculus Mathematics for Calculus

Question

Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. $$y=x^{3}-3 x^{2}, \quad[-2,5] \text { by }[-10,10]$$

Step-by-Step Solution

Answer

The local maximum is at (0, 0) and the local minimum is at (2, -4).

Step 1: Understanding the Polynomial

The given polynomial is \( y = x^3 - 3x^2 \). It is a cubic polynomial, which typically has one or two turning points, known as local extrema (maximum or minimum points). To graph it, we should consider the given region, \([-2, 5]\) by \([-10, 10]\), and start by finding any critical points that would indicate extrema.

Step 2: Find the Derivative

To find the local extrema, we first need the derivative of the polynomial. The derivative is given by taking the derivative of each term: \( y' = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) = 3x^2 - 6x \).

Step 3: Solve Critical Points

Find critical points by setting the derivative to zero: \( 3x^2 - 6x = 0 \). Factor the equation: \( 3x(x - 2) = 0 \), giving critical points at \( x = 0 \) and \( x = 2 \).

Step 4: Determine Local Extrema

To determine if these critical points are maxima or minima, use the second derivative \( y'' = 6x - 6 \). Evaluate: \( y''(0) = -6 \) (indicating a local maximum at \( x = 0 \)), and \( y''(2) = 6 \) (indicating a local minimum at \( x = 2 \)).

Step 5: Compute Extrema Values

Substitute the critical points back into the original polynomial to find the y-values: \( y(0) = 0^3 - 3(0)^2 = 0 \) and \( y(2) = 2^3 - 3(2)^2 = 8 - 12 = -4 \). So, the local maximum is at (0, 0) and the local minimum is at (2, -4).

Step 6: Graph the Polynomial

Plot the points and graph the polynomial in the specified viewing rectangle. Note that the graph should pass through the local extrema points (0, 0) and (2, -4) and follow the cubic shape trends. Verify that the graph falls within \([-10, 10]\) for y-values and \([-2, 5]\) for x-values.

Key Concepts

Critical Points

Critical points are where the first derivative of a function equals zero or is undefined. In simpler terms, these points are potential locations for local maxima or minima on a graph. For the polynomial \( y = x^3 - 3x^2 \), the first derivative is \( y' = 3x^2 - 6x \). By setting this derivative to zero, we find where the slope of the tangent line is horizontal. Solving \( 3x^2 - 6x = 0 \), we factor out \( 3x \), resulting in \( x(x - 2) = 0 \). Thus, the critical points are at \( x = 0 \) and \( x = 2 \).

These x-values indicate where the slope changes, potentially marking the presence of a local extremum. While critical points are necessary candidates for local extrema, further analysis through tests like the second derivative is required to classify them.

Local Extrema

Local extrema refer to the highest or lowest points in a given interval around a critical point. For our polynomial, we found critical points at \( x = 0 \) and \( x = 2 \). To determine whether these points are local maxima or minima, we need to inspect the behavior of the function around these points. We do this with the second derivative test.

Incorporating critical points and testing them is essential in identifying extrema. Remember that a local maximum is the highest point in the neighborhood, whereas a local minimum is the lowest. As the calculation shows in the example, the polynomial has a local maximum at the coordinates (0, 0) and a local minimum at (2, -4). This classification is a crucial step in sketching the polynomial as it guides the curvature and trend of the graph.

Second Derivative Test

The second derivative test is a convenient method for determining if a critical point is a local maximum or minimum. This test involves calculating the second derivative of the function and evaluating it at the critical points.

For our polynomial, the second derivative is \( y'' = 6x - 6 \). At \( x = 0 \), \( y''(0) = 6(0) - 6 = -6 \), which is negative, indicating that the function has a local maximum at this point. On the other hand, for \( x = 2 \), \( y''(2) = 6(2) - 6 = 6 \) is positive, indicating a local minimum at this point.

This test helps quickly determine the nature of critical points. By just looking at the sign of the second derivative, we can immediately conclude whether the function is curving upwards (minimum) or downwards (maximum) at that specific point.