Problem 225 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f \text { , }} \\\ {\text { c. intervals where } f \text { is concave up and concave }} \\\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . }}\end{array} $$ $$ f(x)=x^{3}-6 x^{2} $$

Step-by-Step Solution

Answer

Local max at 0, local min at 4; increasing \((-\infty,0), (4,\infty)\), decreasing \((0,4)\); concave up \((2,\infty)\), concave down \((-\infty,2)\); inflection at 2.

Step 1 - Determine the first derivative

First, differentiate the function \( f(x) = x^3 - 6x^2 \) to find \( f'(x) \). This will help us determine where the function is increasing or decreasing. \[ f'(x) = 3x^2 - 12x \]

Step 2 - Find critical points

Set the first derivative \( f'(x) = 0 \) to find critical points. Solve the equation \( 3x^2 - 12x = 0 \).\[ 3x(x - 4) = 0 \]Therefore, the critical points are \( x = 0 \) and \( x = 4 \).

Step 3 - Test intervals around critical points

Test intervals around the critical points to determine where the function is increasing or decreasing. Use the test points in the intervals \((-\infty, 0)\), \((0, 4)\), and \((4, \infty)\).- For \( x < 0 \), test \( f'(-1) = 3(-1)^2 - 12(-1) = 15 > 0 \) (increasing)- For \( 0 < x < 4 \), test \( f'(2) = 3(2)^2 - 12(2) = -12 < 0 \) (decreasing)- For \( x > 4 \), test \( f'(5) = 3(5)^2 - 12(5) = 15 > 0 \) (increasing)

Step 4 - Determine local minima and maxima

Using the intervals:- The function is decreasing then increasing around \( x = 0 \), so it's a local maximum.- The function is increasing then decreasing around \( x = 4 \), so it's a local minimum.Evaluate the function at these points:\[ f(0) = 0^3 - 6(0)^2 = 0 \]\[ f(4) = 4^3 - 6(4)^2 = 64 - 96 = -32 \]

Step 5 - Determine the second derivative for concavity

Differentiate the first derivative to obtain the second derivative \( f''(x) \). This helps determine intervals of concavity.\[ f''(x) = 6x - 12 \]

Step 6 - Find inflection points

Set the second derivative \( f''(x) = 0 \) to find possible inflection points.\[ 6x - 12 = 0 \]\[ x = 2 \]

Step 7 - Test intervals around inflection points

Test the intervals around the inflection point \( x = 2 \) to determine concavity:- For \( x < 2 \), test \( f''(0) = 6(0) - 12 = -12 < 0 \) (concave down)- For \( x > 2 \), test \( f''(3) = 6(3) - 12 = 6 > 0 \) (concave up)Thus, \( (2, \infty) \) is concave up and \( (-\infty, 2) \) is concave down.

Step 8 - Inflection Points and Summary

The inflection point occurs at \( x = 2 \), as the concavity changes from down to up. **Summary**:- Increasing on \((-\infty, 0)\) and \((4, \infty)\); Decreasing on \((0, 4)\)- Local maximum at \( x = 0 \) and local minimum at \( x = 4 \)- Concave down on \((-\infty, 2)\); Concave up on \((2, \infty)\)- Inflection point at \( x = 2 \)

Key Concepts

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing is fundamental in function analysis. These intervals tell us how the function behaves as you move along the x-axis.
To determine these intervals, we look at the first derivative of the function, denoted as \( f'(x) \), which represents the rate of change. If \( f'(x) > 0 \), then the function is increasing, whereas if \( f'(x) < 0 \), the function is decreasing.
For the function \( f(x) = x^3 - 6x^2 \), the first derivative is \( f'(x) = 3x^2 - 12x \). Setting \( f'(x) = 0 \) gives the critical points, which are \( x = 0 \) and \( x = 4 \). By testing intervals around these critical points:

• The function increases when \( x \) is in intervals \((-\infty, 0)\) and \((4, \infty)\).
• The function decreases in the interval \((0,4)\).

This behavior provides insights into the local trends of the function's graph.

Local Minima and Maxima

Local minima and maxima are points where a function reaches a peak or a trough within a given interval. Analyzing these points helps in understanding the key features of the graph.
In our example, around the critical points \( x=0 \) and \( x=4 \), the nature of the function's increase or decrease changes, indicating potential local extremum points.
When transitioning from increasing to decreasing around a point, a local maximum occurs, whereas a shift from decreasing to increasing suggests a local minimum.
In this case:

• At \( x=0 \), the function transitions from increasing to decreasing, indicating a local maximum with value \( f(0) = 0 \).
• At \( x=4 \), the function transitions from decreasing to increasing, suggesting a local minimum with value \( f(4) = -32 \).

Locating these extremum points helps in plotting and understanding the graph and its strategic high and low points.

Concavity and Inflection Points

Understanding concavity and inflection points requires examining the function's second derivative,\( f''(x) \). Concavity describes the direction and fate of the function graph's bending.
If \( f''(x) > 0 \), the function is concave up (like a cup); if \( f''(x) < 0 \), it is concave down (like a frown).
For the function \( f(x) = x^3 - 6x^2 \), the second derivative is \( f''(x) = 6x - 12 \). Setting \( f''(x) = 0 \) finds inflection points, which mark changes in concavity. In this example:

• \( x = 2 \) is a potential inflection point.
• For \( x < 2 \), \( f''(x) < 0 \) indicating the graph is concave down.
• For \( x > 2 \), \( f''(x) > 0 \) indicating the graph is concave up.

Thus, \( x = 2 \) is an inflection point where the graph switches from concave down to concave up.

Critical Points

Critical points are where a function's first derivative is zero or undefined, providing essential data about the graph's behavior. These points are significant as they indicate potential for local maxima, minima, or horizontality.
For the function \( f(x) = x^3 - 6x^2 \), we find critical points by solving \( f'(x) = 0 \). This calculation leads to:

• \( x = 0 \)
• \( x = 4 \)

These values are checked within the broader context of increasing/decreasing and local extrema. They anchor our understanding of the function's behavior across intervals. Evaluating these points on the original function \( f(x) \) indicates significant values and potential extremum behaviors.
Critical point analysis thus guides the understanding of the overall tendencies and properties of the function graph across specified domains.