Problem 22 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$h(x)=2 x^{3}-18 x$$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -\sqrt{3})\) and \((\sqrt{3}, \infty)\), decreasing on \((-\sqrt{3}, \sqrt{3})\). Local maximum at \(-\sqrt{3}\), local minimum at \(\sqrt{3}\).

Step 1: Find the derivative of the function

To determine where the function is increasing or decreasing, we first need to find the derivative of the function \( h(x) = 2x^3 - 18x \). Using the rules of differentiation, the derivative \( h'(x) \) is \( h'(x) = 6x^2 - 18 \).

Step 2: Determine critical points

Critical points occur where the derivative is zero or undefined. Since \( h'(x) \) is a polynomial, it is defined everywhere. Set the derivative equal to zero: \( 6x^2 - 18 = 0 \). Solve for \( x \) by first factoring: \( 6(x^2 - 3) = 0 \). This simplifies to \( x^2 = 3 \), leading to \( x = \pm\sqrt{3} \). These are your critical points.

Step 3: Test intervals around critical points to determine increase/decrease

Use the critical points to divide the number line into intervals: \((-\infty, -\sqrt{3})\), \((-\sqrt{3}, \sqrt{3})\), and \((\sqrt{3}, \infty)\). Choose a test point in each interval and evaluate \( h'(x) \):- For \( x = -2 \) (in \((-\infty, -\sqrt{3})\)), \( h'(-2) = 6(-2)^2 - 18 = 24 - 18 = 6 \), positive- For \( x = 0 \) (in \((-\sqrt{3}, \sqrt{3})\)), \( h'(0) = 6(0)^2 - 18 = -18 \), negative- For \( x = 2 \) (in \((\sqrt{3}, \infty)\)), \( h'(2) = 6(2)^2 - 18 = 24 - 18 = 6 \), positiveThus, the function is increasing on \( (-\infty, -\sqrt{3}) \) and \((\sqrt{3}, \infty)\), and decreasing on \(( -\sqrt{3}, \sqrt{3}) \).

Step 4: Identify local and absolute extrema

Evaluate the function at the critical points to find potential local extrema: - \( h(-\sqrt{3}) = 2(-\sqrt{3})^3 - 18(-\sqrt{3}) = -6\sqrt{3} + 18\sqrt{3} = 12\sqrt{3} \)- \( h(\sqrt{3}) = 2(\sqrt{3})^3 - 18(\sqrt{3}) = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \) Since the function changes from increasing to decreasing at \(-\sqrt{3}\), this is a local maximum, and where it changes from decreasing to increasing at \(\sqrt{3}\), this is a local minimum. There are no endpoints given to consider absolute extrema.

Key Concepts

Critical Points

Critical points are vital in calculus, as they can tell us about the behavior of the function at certain points. For any function, critical points are found where the function's derivative is either zero or undefined. In our exercise, the derivative of the function, given as \( h'(x) = 6x^2 - 18 \), is always defined because it is a polynomial.

To find the critical points, we set the derivative to zero: \( 6x^2 - 18 = 0 \) and solve for \( x \). By simplifying, we factor it as \( 6(x^2 - 3) = 0 \). Upon solving \( x^2 = 3 \), the solutions are \( x = \pm \sqrt{3} \). These critical points, \( x = -\sqrt{3} \) and \( x = \sqrt{3} \), divide the number line into intervals that we will further examine to understand the function's behavior.

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing helps us get a picture of its overall shape on a graph. To determine whether a function is increasing or decreasing on certain intervals, we use the first derivative test. After finding the critical points, we use them to divide the number line into separate intervals.

• For \( x \) values in \((-\infty, -\sqrt{3})\), the derivative \( h'(-2) = 6 \), indicating the function is increasing in this interval.
• In the interval \((-\sqrt{3}, \sqrt{3})\), the derivative \( h'(0) = -18 \), showing the function is decreasing here.
• Finally, in \((\sqrt{3}, \infty)\), \( h'(2) = 6 \), meaning the function is increasing again.

Thus, our function increases in the intervals \((-\infty, -\sqrt{3})\) and \((\sqrt{3}, \infty)\) while it decreases on the interval \((-\sqrt{3}, \sqrt{3})\). This pattern shows that at \( x = \pm \sqrt{3} \) the function changes direction; these shifts can be key indicators for local extrema.

Local Extrema

Local extrema refer to the highest or lowest points within a certain range of a function. These are the points where a function transitions between increasing and decreasing. At a critical point where a function changes from increasing to decreasing, we find a local maximum, and from decreasing to increasing, a local minimum.

Using our critical points \(x = -\sqrt{3}\) and \( x = \sqrt{3} \):

• At \( x = -\sqrt{3} \), the function changes from increasing to decreasing, indicating a local maximum with a value of \( h(-\sqrt{3}) = 12\sqrt{3} \).
• At \( x = \sqrt{3} \), the function shifts from decreasing to increasing, identifying a local minimum with \( h(\sqrt{3}) = -12\sqrt{3} \).

These points mark significant peaks and troughs in the function and are demonstrated by how the function's slope transitions around these values. No endpoints indicate absolute extrema, so we conclude with these local extrema findings.