Problem 58 - 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. $$ y=x^{5}-5 x^{2}+6, \quad[-3,3] \text { by }[-5,10] $$

Step-by-Step Solution

Answer

Local extrema at approximately (0, 6), (1.26, -0.76), and (-1.26, -0.76).

Step 1: Analyze the Polynomial

The polynomial given is \( y = x^5 - 5x^2 + 6 \). It is a degree 5 polynomial, which means the graph could have up to 4 turning points corresponding to local maxima and minima.

Step 2: Determine Critical Points

To find local extrema, we need to find the critical points by setting the derivative equal to zero. Calculate the derivative of the polynomial: \( y' = 5x^4 - 10x \). Set \( y' = 0 \) to find critical points: \[ 5x^4 - 10x = 0 \] Factor the equation: \[ 5x(x^3 - 2) = 0 \] This gives critical points at \( x = 0 \) and \( x = \sqrt[3]{2} \), including negative root-pair of \( x = -\sqrt[3]{2} \).

Step 3: Determine the Nature of Critical Points

Use the second derivative test to classify the critical points. The second derivative is: \( y'' = 20x^3 - 10 \). Evaluate the second derivative at each critical point. 1. At \( x = 0 \), \( y''(0) = -10 \), indicating a local maximum. 2. At \( x = \sqrt[3]{2} \), calculate \( y''(\sqrt[3]{2}) \), resulting in a positive value, indicating a local minimum. 3. At \( x = -\sqrt[3]{2} \), evaluate \( y''(-\sqrt[3]{2}) \), also giving a positive value, indicating another local minimum.

Step 4: Calculate the Extrema Coordinates

Use the function to calculate the \( y \)-values at the critical points: 1. For \( x = 0 \), \( y = 6 \). 2. For \( x = \sqrt[3]{2} \), substitute back into the original function to find \( y \). 3. For \( x = -\sqrt[3]{2} \), substitute back into the original function to find \( y \). Round all coordinates to two decimal places.

Step 5: Sketch the Graph in Provided Window

Plot the polynomial function over the domain \([-3, 3]\) and range \([-5, 10]\). Ensure to indicate the local extrema found at: \((0, 6)\), \((\sqrt[3]{2}, y(\sqrt[3]{2}))\), and \((-\sqrt[3]{2}, y(-\sqrt[3]{2}))\). Verify visually that these points correspond to local maxima and minima.

Key Concepts

Local Extrema

When graphing polynomials, local extrema are the highest or lowest points in small sections of the graph, excluding endpoints. These points are known as local maxima (highest points) or local minima (lowest points).
Local extrema occur at points where the graph changes direction, creating visible peaks or valleys. To find these points, we search for critical points where the slope of the tangent line is zero.
In the context of our example, we examined the polynomial \( y = x^5 - 5x^2 + 6 \). The task was to identify the specific x-values where the graph had these peaks or valleys in the interval \([-3, 3]\).
Utilizing calculus, we found local minima and maxima by analyzing the critical points where the derivative equals zero.

Derivative

The derivative of a function provides the slope of the tangent line at any given point on its graph. For polynomial graphing, the derivative is crucial in determining the behavior of the polynomial and finding points of local extremum.
To find the derivative, apply basic differentiation rules. In our exercise, the derivative of the polynomial \( y = x^5 - 5x^2 + 6 \) is \( y' = 5x^4 - 10x \). This expression gives the rate of change of \( y \) with respect to \( x \).
Critical points, where potential local extrema exist, occur where this derivative equals zero. By solving \( 5x^4 - 10x = 0 \), we identified the critical points that need further investigation to find whether they are local maxima or minima.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points offer potential for local extrema since they represent where the function's slope changes from positive to negative or vice versa.
For a polynomial, these are solved by setting the derivative equal to zero and factoring the resulting equation. Our given derivative, \( 5x^4 - 10x \), was set to zero resulting in the equation \( 5x(x^3 - 2) = 0 \), leading to critical points at \( x = 0 \), \( x = \sqrt[3]{2} \), and \( x = -\sqrt[3]{2} \).
These critical points were vital in examining where the polynomial could have peaks or valleys, which are interpreted as local maxima and minima in the graph.

Second Derivative Test

The second derivative test is a method to ascertain the nature of critical points in a polynomial function. It helps determine whether these points are local maxima, minima, or saddle points.
By taking the second derivative of the original function, \( y'' = 20x^3 - 10 \) in our exercise, we can evaluate each critical point. The sign of the second derivative at a critical point offers insights:

• If \( y'' > 0 \), the point is a local minimum.
• If \( y'' < 0 \), the point is a local maximum.
• If \( y'' = 0 \), the test is inconclusive.

In our case, evaluating \( y'' \) at \( x = 0 \), \( x = \sqrt[3]{2} \), and \( x = -\sqrt[3]{2} \) provided clear indications of maximum and minimum points crucial for graphing the polynomial accurately.