Problem 11 - Thomas Calculus in SI Units

Question

Answer the following questions about the functions whose derivatives are given in Exercises \(1-14:\) a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=x^{-1 / 3}(x+2)$$

Step-by-Step Solution

Answer

Critical points: \( x = -2 \) and \( x = 0 \). \( f \) decreases on \((-\infty, -2)\), increases on \((-2, \infty)\). Local minimum at \( x = -2 \).

Step 1: Find the Critical Points

The critical points occur where the derivative \( f'(x) = x^{-1/3}(x+2) \) is zero or undefined. The derivative is zero when \( x^{-1/3}(x+2) = 0 \). Solving \( x^{-1/3}(x+2) = 0 \), we find that \( x = -2 \) is a critical point. The derivative is undefined when \( x^{-1/3} \) is undefined, which occurs at \( x = 0 \). Thus, the critical points are \( x = -2 \) and \( x = 0 \).

Step 2: Determine Intervals of Increase or Decrease

We analyze the intervals based on the critical points \( x = -2 \) and \( x = 0 \). Test values in each interval in the derivative: - For \( x < -2 \), pick \( x = -3 \). \( f'(-3) = (-3)^{-1/3}(-3+2) < 0 \) (negative).- For \(-2 < x < 0\), pick \( x = -1 \). \( f'(-1) = (-1)^{-1/3}(1) > 0 \) (positive).- For \( x > 0 \), pick \( x = 1 \). \( f'(1) = (1)^{-1/3}(3) > 0 \) (positive).Thus, \( f \) decreases on \( (-\infty, -2) \), increases on \( (-2, 0) \), and \((0, \infty) \).

Step 3: Find Local Extrema

Examine the critical points for local extrema by checking changes in signs of \( f'(x) \): - At \( x = -2 \), \( f' \) changes from negative to positive. This indicates a local minimum at \( x = -2 \).- At \( x = 0 \), \( f' \) does not change sign; it remains positive. Thus, \( x = 0 \) is not an extremum.

Key Concepts

Derivative Analysis

When working with functions in calculus, the derivative provides extensive information about the function's behavior. The critical points of a function are determined by analyzing its derivative. These points are where the derivative, denoted as \( f'(x) \), is either zero or undefined.

For the function given in the exercise, the derivative is \( f'(x) = x^{-1/3}(x+2) \).

• The derivative equals zero when \( x^{-1/3}(x+2) = 0 \), which occurs at \( x = -2 \).
• The derivative is undefined at \( x = 0 \) since \( x^{-1/3} \) becomes undefined.

Identifying these critical points is the first step in understanding the function's behavior, as they are necessary for determining intervals of increase or decrease and locating local extrema.

Increasing and Decreasing Intervals

Once critical points are identified, the next step is to determine where the function is increasing or decreasing. These intervals tell us how the function behaves between and beyond these critical points.

For the problem, we examine the sign of \( f'(x) \) within intervals divided by the critical points \( x = -2 \) and \( x = 0 \). By picking test points within each interval:

• For \( x < -2 \), \( f'(x) \) is negative, indicating the function is decreasing.
• For \( -2 < x < 0 \), \( f'(x) \) is positive, so the function is increasing.
• For \( x > 0 \), \( f'(x) \) is also positive, meaning the function continues to increase.

Thus, the function decreases on \((-\infty, -2)\) and increases on \((-2, 0)\) and \((0, \infty)\). These intervals assist in sketching the general shape of the function and anticipating its behavior.

Local Extrema

By analyzing the critical points and the intervals of increase and decrease, we can identify the local extrema, which are the points where the function has local maximum or minimum values. These occur at critical points where there is a change in the sign of the derivative \( f'(x) \).

From the problem:

• At \( x = -2 \), the derivative changes from negative to positive, indicating a local minimum at this point.
• At \( x = 0 \), the derivative remains positive without changing signs, so there is no extremum at this point.

Local extrema are significant because they point out the peaks and troughs in the function graph, contributing to the overall understanding of its shape and direction.

Functions and Calculus

Functions and calculus are closely related, serving as fundamental components of mathematical analysis. Understanding how a function behaves requires knowledge of calculus concepts such as derivatives and their applications.

The derivative tells us about the slope of the tangent line to the curve at any given point. This slope information helps us find critical points, intervals of increase and decrease, and any local maxima or minima.

Applying these concepts allows us to effectively analyze and interpret the behavior of functions. By using calculus to find where the derivative is zero or undefined, and then determining the sign changes around those points, we gain valuable insights about a function's graphical representation and real-world applications.