Problem 226 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f \text { , }} \\\ {\text { c. intervals where } f \text { is concave up and concave }} \\\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . }}\end{array} $$ $$ f(x)=x^{4}-6 x^{3} $$

Step-by-Step Solution

Answer

a. Decreasing on \((-\infty, \frac{9}{2})\); increasing on \((\frac{9}{2}, \infty)\). Local minimum at \( x = \frac{9}{2} \). c. Concave up on \((-\infty, 0) \cup (3, \infty)\); concave down on \((0, 3)\). d. Inflection points at \( x = 0 \) and \( x = 3 \).

Step 1: Find the Derivative

To find where the function is increasing or decreasing, calculate the first derivative of the function. The function is \( f(x) = x^4 - 6x^3 \). The first derivative, \( f'(x) \), is obtained by applying the power rule:\[ f'(x) = 4x^3 - 18x^2 \]

Step 2: Solve for Critical Points

Critical points occur where \( f'(x) = 0 \) or is undefined. Solve:\[ 4x^3 - 18x^2 = 0 \]Factor out the common terms:\[ 2x^2 (2x - 9) = 0 \]This gives solutions \( x = 0 \) and \( x = \frac{9}{2} \).

Step 3: Test Intervals Around Critical Points

Test values around the critical points, using the first derivative to determine intervals where \( f \) is increasing or decreasing.- For \( x < 0 \), choose \( x = -1 \): \( f'(-1) = 4(-1)^3 - 18(-1)^2 = -22 \) (decreasing)- For \( 0 < x < \frac{9}{2} \), choose \( x = 1 \): \( f'(1) = 4(1)^3 - 18(1)^2 = -14 \) (decreasing)- For \( x > \frac{9}{2} \), choose \( x = 5 \): \( f'(5) = 4(5)^3 - 18(5)^2 = 50 \) (increasing)

Step 4: Identify Local Minima and Maxima

From the intervals:- The function is decreasing on \((-\infty, 0)\) and \((0, \frac{9}{2})\), then increasing on \((\frac{9}{2}, \infty)\).- At \( x = 0 \), no sign change, no local extrema.- At \( x = \frac{9}{2} \), there is a change from decreasing to increasing, indicating a local minimum.

Step 5: Find Second Derivative for Concavity

To determine concavity, find the second derivative:\[ f''(x) = 12x^2 - 36x \]

Step 6: Solve for Inflection Points

Set the second derivative to zero to find inflection points:\[ 12x^2 - 36x = 0 \]Factor:\[ 12x(x - 3) = 0 \]Solutions are \( x = 0 \) and \( x = 3 \).

Step 7: Test Intervals for Concavity

Use test points around \( x = 0 \) and \( x = 3 \):- For \( x < 0 \), choose \( x = -1 \): \( f''(-1) = 48 \) (concave up)- For \( 0 < x < 3 \), choose \( x = 1 \): \( f''(1) = -24 \) (concave down)- For \( x > 3 \), choose \( x = 4 \): \( f''(4) = 48 \) (concave up)

Step 8: Inflection Points and Concavity Intervals

From the test results:- The function is concave up on \((-fty, 0) \cup (3, fty)\) and concave down on \((0, 3)\).- Inflection points occur at \( x = 0 \) and \( x = 3 \).

Key Concepts

Derivatives

In calculus, the concept of derivatives is crucial for understanding how a function changes at any given point. The derivative of a function provides us with the slope of the tangent line to the curve at a particular point, which indicates the rate of change of the function at that point.
For the function given in the exercise, \( f(x) = x^4 - 6x^3 \), the first step is to find its derivative using the power rule. Applying this rule, we obtain the first derivative as \( f'(x) = 4x^3 - 18x^2 \). This expression helps us understand the function's behavior, particularly where it increases or decreases.
Derivatives are foundational in calculus because they allow us to find critical points and analyze a function's increasing or decreasing behavior.

Critical Points

Critical points of a function are essential for analyzing its local maxima and minima. These points occur where the first derivative \( f'(x) \) is either zero or undefined. Solve \( f'(x) = 0 \) to find these points.
For our function, the critical points are found by solving:

• \( 4x^3 - 18x^2 = 0 \)
• Factor as \( 2x^2(2x - 9) = 0 \)
• This gives solutions \( x = 0 \) and \( x = \frac{9}{2} \).

At these critical points, the behavior of the function can change from increasing to decreasing or vice versa. Understanding critical points allows us to identify potential locations for local extreme values of a function, providing insights into the function's overall shape.

Concavity

Concavity is a property that tells us about the curvature of a function's graph. It indicates whether a function's graph is curving upwards (concave up) or downwards (concave down).
To determine concavity, we compute the second derivative \( f''(x) \). For the function \( f(x) = x^4 - 6x^3 \):

• Calculate \( f''(x) = 12x^2 - 36x \).
• The second derivative helps to analyze the concavity by setting it equal to zero to identify possible inflection points and by testing signs around these points.

Concave up functions look like a U (holding water), and concave down functions look like an inverted U (shedding water), indicating the function's direction around these points.

Inflection Points

Inflection points are points on the graph of a function where concavity changes. They are significant because they show where the function changes its curvature, reflecting a shift from concave up to concave down or vice versa.
To find inflection points:

• Set the second derivative to zero, \( 12x^2 - 36x = 0 \).
• Factor to find \( x = 0 \) and \( x = 3 \).
• Check intervals between and around these solutions to confirm a change in concavity using test points in the second derivative.

This analysis reveals intervals of concavity changes, specifically, the graph of \( f(x) \) is concave up on \((-\infty, 0) \cup (3, \infty)\) and concave down on \((0, 3)\). Identifying inflection points helps in sketching the curve and understanding the shape of the graph.