Problem 22 - Thomas Calculus

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

Increasing on \((-\infty, -\sqrt{3})\) and \((\sqrt{3}, \infty)\); decreasing on \((-\sqrt{3}, \sqrt{3})\). Local max at \(12\sqrt{3}\) and local min at \(-12\sqrt{3}\); no absolute extremum.

Step 1: Find the derivative

To find where the function is increasing or decreasing, first determine the derivative of the function. The derivative, denoted \( h'(x) \), is found using the power rule. For the function \( h(x) = 2x^3 - 18x \), the derivative is calculated as follows: \[ h'(x) = \frac{d}{dx} (2x^3 - 18x) = 6x^2 - 18. \]

Step 2: Find critical points

Critical points occur where the derivative \( h'(x) \) is zero or undefined. Set \( h'(x) = 6x^2 - 18 = 0 \) and solve for \( x \). This gives: \( 6x^2 = 18 \), dividing both sides by 6, we get \( x^2 = 3 \). Taking the square root of both sides gives the critical points \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).

Step 3: Test intervals around critical points

Use the critical points to divide the number line into intervals and test the sign of the derivative \( h'(x) \) in each interval to determine where the function is increasing or decreasing: \[ \text{Intervals: } (-\infty, -\sqrt{3}), (-\sqrt{3}, \sqrt{3}), (\sqrt{3}, \infty) \] Choose test points from each interval:- For \( x = -2 \) (interval \( (-\infty, -\sqrt{3}) \)): \( h'(-2) = 6(-2)^2 - 18 = 6 \times 4 - 18 = 24 - 18 = 6 \) (positive)- For \( x = 0 \) (interval \( (-\sqrt{3}, \sqrt{3}) \)): \( h'(0) = 6(0)^2 - 18 = -18 \) (negative)- For \( x = 2 \) (interval \( (\sqrt{3}, \infty) \)): \( h'(2) = 6(2)^2 - 18 = 6 \times 4 - 18 = 24 - 18 = 6 \) (positive)Thus, \( h(x) \) is increasing on \( (-\infty, -\sqrt{3}) \) and \( (\sqrt{3}, \infty) \), and decreasing on \( (-\sqrt{3}, \sqrt{3}) \).

Step 4: Determine local extrema

Since the derivative changes from positive to negative at \( x = -\sqrt{3} \), there is a local maximum at this point. Similarly, the derivative changes from negative to positive at \( x = \sqrt{3} \), indicating a local minimum. Calculate the function values:- \( h(-\sqrt{3}) = 2(-\sqrt{3})^3 - 18(-\sqrt{3}) = -6\sqrt{3} + 18\sqrt{3} = 12\sqrt{3} \)- \( h(\sqrt{3}) = 2(\sqrt{3})^3 - 18(\sqrt{3}) = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \)Thus, there is a local maximum of \( 12\sqrt{3} \) at \( x = -\sqrt{3} \) and a local minimum of \( -12\sqrt{3} \) at \( x = \sqrt{3} \).

Step 5: Analyze absolute extrema

Since \( h(x) \) is a cubic function which tends to \( \infty \) as \( x \to \infty \) and \( -\infty \) as \( x \to -\infty \), it does not have any absolute maximum or minimum. The local extrema determined are not absolute because the function continues to grow beyond these points.

Key Concepts

Derivative

The derivative of a function gives us critical information about the behavior of the function at different points. It tells us how the function is changing — whether it is increasing or decreasing as we move along the graph. For our function, which is \( h(x) = 2x^3 - 18x \), the derivative is calculated using the power rule. By using the power rule, each term's exponent is decreased by one after being multiplied by the original exponent value. Here, this gives us:

• From \( 2x^3 \), multiplying 3 by 2 and subtracting 1 from the exponent gives us \( 6x^2 \).
• For \( -18x \), we multiply 1 by -18, resulting in \( -18 \) (since the exponent is reduced by one, which becomes zero, making it a constant).

So, the derivative is \( h'(x) = 6x^2 - 18 \). Finding where this derivative is zero or undefined helps us discover critical points where the function's slope changes, indicating potential local maxima or minima.

Increasing and Decreasing Intervals

By understanding the derivative, we identify how a function behaves across different intervals. Increasing intervals mean the function goes upwards as \( x \) increases, while decreasing intervals suggest the opposite. With \( h'(x) = 6x^2 - 18 \), we set the derivative equal to zero to find critical points:

• \( 6x^2 - 18 = 0 \)
• Solving this, \( x^2 = 3 \), leads to critical points at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).

By testing points around these critical values, such as \( x = -2, 0, \text{and } 2 \), we determine the sign of the derivative in each interval and hence whether the function is increasing or decreasing. For \( x = -2 \) and \( x = 2 \), \( h'(x) \) is positive, indicating increasing intervals \((-\infty, -\sqrt{3}) \) and \((\sqrt{3}, \infty) \). At \( x = 0 \), \( h'(x) \) is negative, showing a decreasing interval \((-\sqrt{3}, \sqrt{3}) \). This method efficiently segments the graph into regions of increase and decrease.

Local Extrema

Local extrema refer to the local highest or lowest points within a particular interval on the graph of a function. They occur at critical points where the function changes from increasing to decreasing or vice versa. In the case of our function, the derivative changes:

• From positive to negative at \( x = -\sqrt{3} \), indicating a local maximum.
• From negative to positive at \( x = \sqrt{3} \), indicating a local minimum.

Calculating the actual function values at these points provides us:

• At \( x = -\sqrt{3} \), \( h(-\sqrt{3}) = 12\sqrt{3} \) is a local maximum.
• At \( x = \sqrt{3} \), \( h(\sqrt{3}) = -12\sqrt{3} \) is a local minimum.

This analysis provides insight into the behavior of the function at specific points, allowing better understanding of its overall shape.

Cubic Functions

Cubic functions, such as \( h(x) = 2x^3 - 18x \), are polynomial functions of degree three. They form unique curves that can exhibit up to two turns or bends. Cubic functions do not have horizontal asymptotes, but they extend to infinity or negative infinity and can have local extrema without having absolute maxima or minima. The end behavior of cubic functions is significant:

• As \( x \to \infty \), \( h(x) \to \infty \).
• As \( x \to -\infty \), \( h(x) \to -\infty \).

This means the function continues to rise or fall without bounds in both directions. Understanding cubic functions is critical for grasping more complex polynomial behavior in calculus. They often provide simpler models for real-world situations where changes in direction occur, highlighting their role in both mathematical concepts and practical applications.