Problem 121 - Thomas Calculus

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? \begin{equation}y=x^{5}-5 x^{4}-240\end{equation}

Step-by-Step Solution

Answer

Inflection points: \(x = 0, 3\). Local max at \((0, -240)\), local min at \((4, -496)\).

Step 1: Find the First Derivative

Calculate the first derivative of the function \(y = x^5 - 5x^4 - 240\) to analyze local maxima and minima. We use the power rule:\[y' = \frac{d}{dx}(x^5 - 5x^4 - 240) = 5x^4 - 20x^3.\]

Step 2: Find the Critical Points

Set the first derivative equal to zero to find the critical points:\[5x^4 - 20x^3 = 0.\]Factor out the greatest common factor:\[5x^3(x - 4) = 0.\]Thus, the potential critical points are \(x = 0\) and \(x = 4\).

Step 3: Find the Second Derivative

Calculate the second derivative to determine concavity and inflection points:\[y'' = \frac{d}{dx}(5x^4 - 20x^3) = 20x^3 - 60x^2.\]

Step 4: Find Inflection Points

Set the second derivative equal to zero to find inflection points:\[20x^3 - 60x^2 = 0.\]Factor out the greatest common factor:\[20x^2(x - 3) = 0.\]Thus, the points where potential inflections occur are \(x = 0\) and \(x = 3\).

Step 5: Test Concavity with the Second Derivative

Use the second derivative test for concavity at critical points. Evaluate \(y''(x)\) at points slightly less than and greater than \(x = 0\) and \(x = 4\):- For \(x = 0\): \(y''(-1) > 0\) and \(y''(1) < 0\) (indicating a possible local maximum at \(x = 0\)).- For \(x = 4\): \(y''(3.9) < 0\) and \(y''(4.1) > 0\) (indicating a possible local minimum at \(x = 4\)).

Step 6: Verify Function Values of Critical Points

Calculate the function values at critical points to find exact maximum and minimum coordinates:- For \(x = 0\), \(y(0) = 0^5 - 5 \times 0^4 - 240 = -240\).- For \(x = 4\), \(y(4) = 4^5 - 5 \times 4^4 - 240 = -496.\)The local maximum point is \((0, -240)\) and local minimum is \((4, -496)\).

Step 7: Analyze the Relationship of Derivative Graphs

The intersection points of the first derivative graph \(y' = 5x^4 - 20x^3\) with the \(x\)-axis represent where the function's slope is zero, indicating possible maxima and minima. The second derivative graph \(y'' = 20x^3 - 60x^2\) intersects the \(x\)-axis at potential inflection points where concavity changes.

Key Concepts

Inflection Points

Inflection points occur where the graph of a function changes its concavity, that is, where the curve shifts from concave up to concave down, or vice versa. To determine these points, we use the second derivative of a function. The second derivative generally tells us about the curvature of the function. If we set the second derivative equal to zero and solve for the variable, we get potential inflection points.

• This process involves solving the equation obtained by setting the second derivative to zero: \( 20x^3 - 60x^2 = 0 \).
• By factoring out the greatest common factor, we can find the potential inflection points at \( x = 0 \) and \( x = 3 \).
• To confirm whether these points are true inflection points, check the sign of the second derivative before and after these points to ensure a change in concavity.

This change of concavity indicates that these are valid inflection points on the graph.

Local Maximum and Minimum

Local maxima and minima, also known as extrema, are the highest and lowest points in a particular region of a function. At these points, the slope of the tangent to the function is zero, meaning the first derivative of the function is zero.

• The procedure to find local maxima and minima involves setting the first derivative equal to zero: \( 5x^4 - 20x^3 = 0 \).
• After factoring, we find critical points at \( x = 0 \) and \( x = 4 \).
• To determine whether these critical points are maxima or minima, use the second derivative test. If the second derivative at the critical point is positive, the point is a local minimum. If it's negative, the point is a local maximum.

In this example, \( x = 0 \) results in a local maximum value of \( y = -240 \), and \( x = 4 \) provides a local minimum value of \( y = -496 \).

First Derivative

The first derivative of a function gives us crucial information about the function's increasing or decreasing behavior. In other words, it helps identify slopes and potential local maxima or minima.

The first derivative of the given function \( y = x^5 - 5x^4 - 240 \) is:

• \( y' = 5x^4 - 20x^3 \).
• Critical points occur where \( y' = 0 \).

To solve \( 5x^4 - 20x^3 = 0 \), factor out the greatest common factor to find \( x = 0 \) and \( x = 4 \) as critical points. These points are where the slope changes from increasing to decreasing or vice versa, indicating potential maxima or minima. By analyzing these slopes, we can comprehend how the function behaves across different intervals.

Second Derivative

The second derivative of a function provides insights into the concavity of the graph. It helps us determine whether the graph is curving upwards or downwards, which is key for finding inflection points and further confirming local maxima and minima.

The second derivative of our function is:

• \( y'' = 20x^3 - 60x^2 \).
• We set this equal to zero to find potential inflection points.

Solving \( 20x^3 - 60x^2 = 0 \), by factoring out \( 20x^2 \), results in \( x = 0 \) and \( x = 3 \) as tentative inflection points. To conclude whether they are valid, examine sign changes of \( y'' \) around these points. Additionally, using the second derivative test at critical points, we can confirm whether a critical point is a local maximum (\( y'' < 0 \)) or a local minimum (\( y'' > 0 \)).