Problem 8 - Thomas Calculus

Question

Answer the following questions about the functions whose derivatives are given in Exercises \(1-8 :\) a. What are the critical points of \(f ?\) b. On what intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$ f^{\prime}(x)=x^{-1 / 2}(x-3) $$

Step-by-Step Solution

Answer

Critical points are at \(x = 0\) and \(x = 3\), with \(f\) decreasing on \((0, 3)\) and increasing on \((3, \infty)\). Local minimum at \(x = 3\).

Step 1: Find the critical points

Critical points occur where the derivative \(f'(x)\) is zero or undefined. The given derivative is \(f'(x) = x^{-1/2}(x-3)\). First, find where it is undefined: set the denominator of \(x^{-1/2}\), which is \(x^{1/2}\), equal to zero. This occurs at \(x = 0\). For where \(f'(x) = 0\), set the numerator \(x-3 = 0\), resolving to \(x = 3\). Hence, the critical points are \(x = 0\) and \(x = 3\).

Step 2: Determine intervals of increase and decrease

To determine where \(f\) is increasing or decreasing, examine the sign of \(f'(x)\). Test intervals around the critical points: **Interval 1:** \((-\infty, 0),\) **Interval 2:** \((0, 3),\) and **Interval 3:** \((3, \infty)\). For \(x < 0\), \(x^{-1/2}\) is undefined, hence cannot test. For \(x = 1\) (in Interval 2), \(f'(1) = 1^{-1/2}(1-3) < 0\), so \(f\) is decreasing. For \(x = 4\) (in Interval 3), \(f'(4) = 4^{-1/2}(4-3) > 0\), so \(f\) is increasing.

Step 3: Identify local maxima and minima

The function decreases up to \(x = 3\) and begins to increase thereafter, indicating a local minimum at \(x = 3\). Since \(f'(x)\) is undefined at \(x = 0\) and the direction changes from undefined directly to decreasing, \(x = 0\) is not a minimum or maximum within a neighborhood of defined \(f'(x)\).

Key Concepts

Increasing and Decreasing Intervals

Understanding increasing and decreasing intervals of a function is useful for analyzing how the function behaves over different sections of its domain. To determine these intervals, we use the sign of the derivative, in this case, denoted as \(f'(x)\), which tells us how the function \(f(x)\) is changing.
First, identify the critical points of \(f\) by setting the derivative \(f'(x) = x^{-1/2}(x-3)\) to zero or finding where it is undefined. These are \(x = 0\) and \(x = 3\). These points divide the x-axis into intervals:

• Interval 1: \((-\infty, 0)\)
• Interval 2: \((0, 3)\)
• Interval 3: \((3, \infty)\)

For interval analysis, skip \(x < 0\) as \(f'(x)\) is undefined. Test a value in each remaining interval:

• In Interval 2, test \(x = 1\). The derivative is negative, so \(f(x)\) is decreasing.
• In Interval 3, test \(x = 4\). The derivative is positive, so \(f(x)\) is increasing.

By examining the sign of \(f'(x)\) in each interval, we can conclude where \(f(x)\) is increasing or decreasing.

Local Maximum and Minimum Values

Local maximum and minimum values provide insight into the highest or lowest point within a specified interval of the function. They are found using the critical points and the behavior of \(f(x)\) around these points.
From our analysis of intervals:

• At \(x = 3\), the function changes from decreasing (Interval 2) to increasing (Interval 3), indicating a local minimum.

This transition point, \(x = 3\), is crucial; it shows a bottoming out of \(f(x)\) as it stops decreasing and starts increasing.
Knowing this, we recognize that \(x = 3\) is a local minimum where the function \(f\) bottoms out before rising again. It's important to note that at \(x=0\), \(f'(x)\) is undefined, and thus \(x=0\) is not considered a local maximum or minimum.

Derivative Analysis

Exploring how derivatives are used in function analysis helps us deeply understand its behavior and characteristics. The process generally involves finding and examining the derivative's sign to gain insights into the function \(f(x)\).
The derivative \(f'(x) = x^{-1/2}(x-3)\) contains two factors: \(x^{-1/2}\) and \(x-3\), each influencing the sign of the derivative.

• \(x^{-1/2}\) becomes undefined at \(x=0\) and changes from undefined to real and positive as \(x\) increases beyond zero.
• The term \(x-3\) changes from negative to positive at \(x=3\).

Together, these factors determine \(f'(x)'s\) sign:

• For \(x > 3\), both components are positive making \(f'(x)\) positive and \(f(x)\) increasing.
• For \(x\) in the interval \((0, 3)\), \(x-3\) is negative while \(x^{-1/2}\) remains positive, meaning \(f'(x)\) is negative and \(f(x)\) is decreasing.

Understanding these shifts aids in comprehensively analyzing the function's behavior and determining critical points for further exploration.