Problem 114 - 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? $$y=x^{3}-12 x^{2}$$

Step-by-Step Solution

Answer

Local max at (0,0); local min at (8,64); inflection at (4,-64).

Step 1: Find the first derivative

To find potential local maxima and minima, we need to determine the critical points by setting the first derivative equal to zero. For the function \( y = x^3 - 12x^2 \), find the first derivative with respect to \( x \):\[ y' = \frac{d}{dx}(x^3 - 12x^2) = 3x^2 - 24x \].

Step 2: Solve for critical points

Set the first derivative equal to zero to find the critical points:\[ 3x^2 - 24x = 0 \]Factor the equation:\[ 3x(x - 8) = 0 \].Thus, the critical points are \( x = 0 \) and \( x = 8 \).

Step 3: Use the second derivative test

The second derivative test will help determine if the critical points are local minima or maxima. First, find the second derivative of the function:\[ y'' = \frac{d}{dx}(3x^2 - 24x) = 6x - 24 \].

Step 4: Evaluate the second derivative at the critical points

Substitute the critical points into the second derivative:- At \( x = 0 \), \( y'' = 6(0) - 24 = -24 \), indicating a local maximum since \( y'' < 0 \).- At \( x = 8 \), \( y'' = 6(8) - 24 = 24 \), indicating a local minimum since \( y'' > 0 \).

Step 5: Find the inflection point

An inflection point occurs where the second derivative is zero and changes sign. Solve \( y'' = 0 \):\[ 6x - 24 = 0 \]\[ x = 4 \].

Step 6: Determine the y-coordinates of critical and inflection points

Substitute \( x = 0 \), \( x = 8 \), and \( x = 4 \) back into the original function \( y = x^3 - 12x^2 \):- At \( x = 0 \), \( y = 0^3 - 12 \, 0^2 = 0 \).- At \( x = 8 \), \( y = 8^3 - 12 \, 8^2 = 64 \).- At \( x = 4 \), \( y = 4^3 - 12 \, 4^2 = -64 \).Thus, the coordinates are (0,0) for the local maximum, (8,64) for the local minimum, and (4,-64) for the inflection point.

Step 7: Graph the function and its derivatives

Plot the original function \( y = x^3 - 12x^2 \), and its first derivative \( y' = 3x^2 - 24x \), and second derivative \( y'' = 6x - 24 \). The graph should clearly show:- The function\( y \) crosses the \( x \)-axis at the critical points (0,0) and (8,64).- The first derivative \( y' \) crossing the \( x \)-axis at the critical points (0,0) and (8,0), indicating potential maxima and minima.- The second derivative \( y'' \) crossing the \( x \)-axis at \( x=4 \) indicates the inflection point. The first derivative changes sign around critical points, and the second derivative changes sign around the inflection point. This establishes the nature of the local maximum, minimum, and inflection.

Key Concepts

Local Maximum and Minimum

To find the local maximum and minimum of a function, we evaluate the function's critical points—where the first derivative equals zero or is undefined. For instance, in the function given by \( y = x^3 - 12 x^2 \), the critical points identified were \( x = 0 \) and \( x = 8 \).
At these points, we utilize the second derivative test to determine the behavior of the function:

• A positive value of the second derivative at a critical point suggests a local minimum.
• A negative value indicates a local maximum.

For example, at \( x = 0 \), since the second derivative \( y'' = -24 \) (less than 0), the graph shows a local maximum at \((0,0)\). Conversely, at \( x = 8 \), with \( y'' = 24 \) (greater than 0), there is a local minimum at \((8,64)\). Understanding these points helps visualize where the graph has upward or downward peaks.

First Derivative

The first derivative of a function, often denoted as \( y' \), provides insights into its increasing or decreasing nature. By calculating \( y' \), we can identify critical points which potentially indicate where local maximum or minimum points occur.
For the function \( y = x^3 - 12x^2 \), the first derivative is \( y' = 3x^2 - 24x \). Setting this equal to zero solved:

• Factor to find: \( 3x(x - 8) = 0 \).
• Resulting critical points: \( x = 0 \) and \( x = 8 \).

This reveals where the slope of the tangent line is horizontal. The first derivative graph intersects the \( x \)-axis at these critical points, suggesting where potential maxima or minima are located within the original graph.

Second Derivative

The second derivative, represented as \( y'' \), tells us about the concavity of a graph—whether it is curved upwards or downwards. This is crucial in determining the inflection points and also aids in the second derivative test to identify local maxima and minima.
For \( y = x^3 - 12x^2 \), the second derivative is calculated as \( y'' = 6x - 24 \). It is used to:

• Evaluate the nature of critical points: \( y'' < 0 \) for maxima and \( y'' > 0 \) for minima.
• Identify inflection points by finding where \( y'' = 0 \). Here, \( x = 4 \).

Understanding this helps us realize where the function changes its curving direction, which occurs at \( (4,-64) \), elucidating the shift from concave up to concave down or vice versa.

Critical Points

Critical points are where the first derivative is zero or undefined and play a pivotal role in understanding the potential turning points on a graph. For the function \( y = x^3 - 12x^2 \), we found these critical points by setting the first derivative \( y' = 3x^2 - 24x \) to zero.
Subsequently:

• Solve \( 3x(x - 8) = 0 \) to get \( x = 0 \) and \( x = 8 \).

These x-values help determine where we have potential peaks (local maxima) or troughs (local minima) on the graph.
After calculating the y-values, the critical points help us determine:

• A local maximum at \( (0,0) \)
• A local minimum at \( (8,64) \)

Understanding critical points allows us to sketch accurate graphs and contextualize features like maximas, minimas, and inflection points effectively.