Problem 113 - Thomas Calculus in SI Units

Question

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\) -axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? $$y=x^{5}-5 x^{4}-240$$

Step-by-Step Solution

Answer

Local minimum at (4, -496); inflection points at x = 0 and x = 3.

Step 1: Find First Derivative

To find critical points and determine where the function has local maximum or minimum values, calculate the first derivative of the given function. The function is \( y = x^5 - 5x^4 - 240 \). Use the power rule to differentiate: \[ y' = \frac{d}{dx}(x^5) - \frac{d}{dx}(5x^4) \]\[ y' = 5x^4 - 20x^3 \].

Step 2: Critical Points of First Derivative

Set the first derivative equal to zero to find critical points:\[ 5x^4 - 20x^3 = 0 \]Factor out the common term, which is \(5x^3\):\[ 5x^3(x - 4) = 0 \]This gives critical points at \( x = 0 \) and \( x = 4 \).

Step 3: Test for Local Maxima and Minima

Use the second derivative test to determine the nature of each critical point. First, calculate the second derivative:\[ y'' = \frac{d}{dx}(5x^4 - 20x^3) \]\[ y'' = 20x^3 - 60x^2 \].Evaluate the second derivative at each critical point. - At \( x = 0 \): \[ y''(0) = 20(0)^3 - 60(0)^2 = 0 \] Inflection point possible, check further if required.- At \( x = 4 \): \[ y''(4) = 20(4)^3 - 60(4)^2 = 1280 - 960 = 320 \] Since \( y''(4) > 0 \), there is a local minimum at \( x = 4 \).

Step 4: Solve for Inflection Points

To find inflection points, set the second derivative equal to zero and solve for \(x\):\[ 20x^3 - 60x^2 = 0 \]Factor out common terms:\[ 20x^2(x - 3) = 0 \]Solve for \(x\) to get possible inflection points at \( x = 0 \) and \( x = 3 \).

Step 5: Substitute to Find Coordinates of Critical Points

Substitute \( x = 0 \) and \( x = 4 \) into the original function to find the coordinates of these points:- For \( y(0) = 0^5 - 5(0)^4 - 240 = -240 \).- For \( y(4) = 4^5 - 5(4)^4 - 240 = 1024 - 1280 - 240 = -496 \).Thus, the points are (0, -240) and (4, -496).

Step 6: Graph Function and Derivatives

Using a graphing tool, plot: - The function \( y = x^5 - 5x^4 - 240 \).- Its first derivative \( y' = 5x^4 - 20x^3 \).- Its second derivative \( y'' = 20x^3 - 60x^2 \).Ensure the graph displays behaviors at the critical points and inflection points identified. The points at which each derivative graph intersects the x-axis are related to behavior changes in the function graph.

Step 7: Conclusion: Relation of Intersections and Derivatives

The function has a local minimum at \((4, -496)\). The first derivative intersects the x-axis at the critical points, and the second derivative intersects at points that may indicate inflection points (\(x = 0\) and \(x = 3\)). These derivatives help confirm critical point types and curvature changes.

Key Concepts

First Derivative Test

The First Derivative Test is a method used to determine local maxima or minima of a function. This involves calculating the first derivative of a function using the power rule, setting it equal to zero, and solving for the critical points. These critical points are where the slope of the tangent to the curve is zero, indicating potential peaks or valleys.

For example, consider the function \( y = x^5 - 5x^4 - 240 \). The first derivative, obtained via the power rule, is \( y' = 5x^4 - 20x^3 \). Set this derivative equal to zero: \( 5x^4 - 20x^3 = 0 \). Factor out common terms to solve for \( x \), leading to critical points at \( x = 0 \) and \( x = 4 \).

• At these critical points, the function might reach a local maximum or minimum, depending on further analysis using the First Derivative Test.

We use the First Derivative Test by investigating the sign changes of \( y' \) around these critical points. If \( y' \) changes from positive to negative, the point is a local maximum; if from negative to positive, it's a local minimum.

Second Derivative Test

The Second Derivative Test helps confirm the nature of critical points found using the First Derivative Test. It involves computing the second derivative of a function. If the second derivative is greater than zero at a critical point, the function has a local minimum there. If it's less than zero, the function has a local maximum.

In our example, calculate the second derivative: \( y'' = 20x^3 - 60x^2 \). Evaluate \( y'' \) at each critical point.

• At \( x = 0 \), \( y''(0) = 0 \). This isn't conclusive, so further testing, such as the First Derivative Test or checking the change of concavity, is needed.
• At \( x = 4 \), \( y''(4) = 320 \), indicating a local minimum at \( x = 4 \) since the second derivative is positive.

The Second Derivative Test offers a straightforward method to confirm the nature of extrema identified through the first derivative test.

Power Rule

The Power Rule is a basic technique in calculus for differentiating functions involving powers of \( x \). It states that for a function \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \). This is crucial for finding derivatives quickly and efficiently.

For example, for the function \( y = x^5 - 5x^4 - 240 \), apply the power rule:

• Derivative of \( x^5 \) is \( 5x^{5-1} = 5x^4 \).
• Derivative of \( -5x^4 \) is \(-20x^{4-1} = -20x^3 \).

Combined, these give \( y' = 5x^4 - 20x^3 \).
The power rule is one of the first tools learned in calculus and is essential for understanding more complex differentiation.

Graphing Derivatives

Graphing derivatives is an insightful way to understand the behavior of a function and its rate of change. The first derivative graph shows where the original function is increasing or decreasing, while the second derivative provides information about concavity and inflection points.

Plotting the first derivative \( y' = 5x^4 - 20x^3 \), notice where it intersects the x-axis. These intersections, at \( x = 0 \) and \( x = 4 \), correspond to critical points in the original function.

• Where the graph is above the x-axis, the function is increasing.
• Where it's below, the function is decreasing.

The second derivative \( y'' = 20x^3 - 60x^2 \), when graphed, reveals where the function changes concavity. Points where this second derivative graph intersects the x-axis can indicate potential inflection points, like at \( x = 0 \) and \( x = 3 \).
Graphing derivatives not only confirms analytical findings but also provides a visual understanding of a function’s dynamics, aiding deeper comprehension.