Chapter 4: Problem 54
Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer correct to two decimal places. $$ y=x^{5}-5 x^{2}+6, \quad[-3,3] \text { by }[-5,10] $$
Short Answer
Expert verified
Local max at (0, 6) and local min at (~1.26, ~1.37).
Step by step solution
01
Understand the Problem
We need to analyze the polynomial \( y = x^5 - 5x^2 + 6 \) and identify its local extrema within the specified viewing rectangle, [-3, 3] by [-5, 10].
02
Find the Derivative
Calculate the first derivative of the polynomial to find the critical points. The first derivative is \( y' = 5x^4 - 10x \).
03
Solve for Critical Points
Set the first derivative equal to zero to find critical points: \( 5x^4 - 10x = 0 \). Factor this expression: \( 5x(x^3 - 2) = 0 \). The solutions are \( x = 0 \) and \( x = \sqrt[3]{2} \), noting only real roots matter in this context.
04
Evaluate the Second Derivative
Calculate the second derivative as it helps in determining the nature of these critical points: \( y'' = 20x^3 - 10 \).
05
Determine Nature of Critical Points
Evaluate the second derivative at \( x = 0 \) and \( x = \sqrt[3]{2} \). For \( x = 0 \), \( y''(0) = -10 \) (indicating a local maximum). For \( x = \sqrt[3]{2} \), \( y''(\sqrt[3]{2}) = 10 > 0 \), indicating a local minimum.
06
Calculate y-Values of Extrema
Substitute back into the original polynomial to find the \( y \)-values. For \( x = 0 \), \( y(0) = 6 \). For \( x = \sqrt[3]{2} \), compute \( y(\sqrt[3]{2}) = (\sqrt[3]{2})^5 - 5(\sqrt[3]{2})^2 + 6 \).
07
Graph the Polynomial (Consider the Viewport)
Sketch the graph using the polynomial equation on a grid defined by \( x \) from -3 to 3 and \( y \) from -5 to 10. Highlight and mark the local extrema points \( (0, 6) \) and \( (\sqrt[3]{2}, y(\sqrt[3]{2})) \).
08
Calculate Approximate Coordinate of Local Minimum to Two Decimal Places
Approximating \( \sqrt[3]{2} \), we find it's approximately 1.26. Evaluate \( y(1.26) = 1.26^5 - 5(1.26)^2 + 6 \) to get \( y \) value at the local minimum. The calculated value is approximately 1.37.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Local Extrema
Local extrema of a polynomial are the high and low points on its graph, which can be either a local maximum or minimum. In simpler terms, these are the turning points where the function changes direction.
For the polynomial given, we are tasked to find these points within a certain range.
In our exercise, the local maximum was found at \( x = 0 \) with a value of 6, and a local minimum was approximated at \( x \approx 1.26 \), with the \( y \)-value approximately 1.37.
For the polynomial given, we are tasked to find these points within a certain range.
- A **local maximum** occurs where the function changes from increasing to decreasing.
- A **local minimum** is found where the function changes from decreasing to increasing.
In our exercise, the local maximum was found at \( x = 0 \) with a value of 6, and a local minimum was approximated at \( x \approx 1.26 \), with the \( y \)-value approximately 1.37.
Derivative
The concept of a derivative is vital in understanding how a function behaves. In the context of graphing polynomials, the derivative tells us the slope or rate of change of the function at any point.
For this exercise:
For this exercise:
- The first derivative, given by \( y' = 5x^4 - 10x \), represents the slope of the polynomial at each \( x \).
- By setting the first derivative to zero, \( y' = 0 \), we find **critical points** where the slope is zero, indicating potential local extrema.
Second Derivative
The second derivative of a polynomial provides information about the curvature or concavity of the function. When graphing polynomials, knowing the second derivative helps us understand whether the graph is curving upwards or downwards at a critical point.
The second derivative \( y'' = 20x^3 - 10 \) was computed. Evaluating this at the critical points allowed us to determine the nature of the extrema:
- If the second derivative \( y'' \) is positive at a particular \( x \), it indicates that the graph is concave up, pointing to a local minimum.
- If \( y'' \) is negative, the graph is concave down, indicating a local maximum.
The second derivative \( y'' = 20x^3 - 10 \) was computed. Evaluating this at the critical points allowed us to determine the nature of the extrema:
- At \( x = 0 \), \( y''(0) = -10 \) signifying a local maximum.
- At \( x = \sqrt[3]{2} \), \( y''(\sqrt[3]{2}) \approx 10 \), confirming a local minimum.
Critical Points
Critical points are the backbone of finding local extrema in a polynomial's graph. These are the points where the first derivative is zero or undefined, suggesting possible areas where the graph might turn.
After the calculations in our exercise:
- To find these points, solve \( 5x^4 - 10x = 0 \) for \( x \).
- This can be factored to \( 5x(x^3 - 2) = 0 \), giving solutions \( x = 0 \) and \( x = \sqrt[3]{2} \).
After the calculations in our exercise:
- At \( x = 0 \), a local maximum was identified.
- Near \( x = 1.26 \), a local minimum was approximated for this polynomial.