Problem 55 - Algebra and Trigonometry

Question

\(51-58\) Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. $$ y=x^{4}+4 x^{3}, \quad[-5,5] \text { by }[-30,30] $$

Step-by-Step Solution

Answer

Local minimum at \((-3, -27)\). No clear local maximum; inflection at \((0,0)\).

Step 1: Understand the Function

We are given the polynomial function \( y = x^4 + 4x^3 \). The task is to graph this function within the specified viewing rectangle, \([-5, 5]\) for the \(x\)-axis and \([-30, 30]\) for the \(y\)-axis, and determine any local extrema.

Step 2: Derivative to Find Critical Points

To find the local extrema, we first need to take the derivative of \( y \) with respect to \( x \). The derivative is:\[y' = \frac{d}{dx}(x^4 + 4x^3) = 4x^3 + 12x^2.\]Set the derivative equal to zero to find the critical points:\[4x^3 + 12x^2 = 0.\]

Step 3: Solve for Critical Points

Factor the derivative equation:\[4x^2(x + 3) = 0.\]This gives the critical points at \( x = 0 \) and \( x = -3 \). These are the potential local extrema.

Step 4: Second Derivative Test

To determine the nature of these critical points, we take the second derivative:\[y'' = 12x^2 + 24x.\]Evaluate the second derivative at the critical points:- At \( x = 0 \), \( y''(0) = 0 \) (test further or use graphical analysis).- At \( x = -3 \), \( y''(-3) = 108 \) (since positive, local minimum).

Step 5: Find Function Values at Critical Points

Calculate the function values at the critical points to find the coordinates:- At \( x = 0 \), \( y(0) = 0^4 + 4(0)^3 = 0 \).- At \( x = -3 \), \( y(-3) = (-3)^4 + 4(-3)^3 = 81 - 108 = -27 \).

Step 6: Plot and Analyze Graph

Graph the function \( y = x^4 + 4x^3 \) using the provided viewing rectangle to observe the behavior of the polynomial. The graph should show a local minimum at \( x = -3 \) and \( y = -27 \). The point \( (0,0) \) may appear as a point of inflection rather than a local extremum.

Key Concepts

Critical Points

To understand critical points, let's begin by defining them. Critical points of a function are where the derivative \( y' \) is zero or undefined. These points are potential locations for local maxima or minima because they indicate where the function's slope changes signs. In our exercise, we have the function \( y = x^4 + 4x^3 \), and its derivative is \( y' = 4x^3 + 12x^2 \).

To find critical points, we set the derivative equal to zero:

• \( 4x^3 + 12x^2 = 0 \)
• This equation factors to \( 4x^2(x + 3) = 0 \)
• Solving gives \( x = 0 \) and \( x = -3 \)

These are our critical points. Once found, these points help us determine where the function might have peaks or valleys. Observing these changes is crucial in graphing and analyzing functions.

Local Extrema

Local extrema refer to the peaks and valleys of a function where a change in direction occurs. Specifically, it's where a function reaches a local maximum or minimum value within a specific region rather than for the entire domain. Identifying such points involves finding where the derivative equals zero and applying further tests to classify them.

In our exercise, the critical points \( x = 0 \) and \( x = -3 \) are already determined. To find the local extrema coordinates:

• Evaluate the function at these critical points.
• For \( x = 0 \), the function value is \( y(0) = 0 \).
• For \( x = -3 \), the function value is \( y(-3) = -27 \).

This analysis shows \( (-3, -27) \) as a local minimum. Since \( (0, 0) \) did not result in a local minimum or maximum from further testing, it might be a point of inflection instead. This analysis helps in understanding the function's layout effectively.

Second Derivative Test

The Second Derivative Test is a method used to determine the nature of critical points found from the first derivative. By checking if the second derivative at a critical point is positive, negative, or zero, you can classify it as a local minimum, maximum, or neither, respectively. It provides a more profound insight into the shape of the graph at those critical points.

In this exercise, the second derivative of the function \( y = x^4 + 4x^3 \) is found to be \( y'' = 12x^2 + 24x \). We evaluate this at our critical points:

• For \( x = 0 \), \( y''(0) = 0 \) indicates further testing is necessary, or a graph is required to conclude.
• For \( x = -3 \), \( y''(-3) = 108 \), which is positive, indicating a local minimum.

Thus, through the second derivative test, we confirm that \( x = -3 \) is indeed a local minimum point. Using this test allows for a comprehensive understanding of the graph's behavior at its critical points, solidifying the categorization of the extrema.