Problem 12 - Thomas Calculus

Question

Answer the following questions about the functions whose derivatives are given. 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? $$ f^{\prime}(x)=x^{-1 / 2}(x-3) $$

Step-by-Step Solution

Answer

Critical points at x = 0 and x = 3; f is decreasing on (0, 3) and increasing on (3, ∞); local minimum at x = 3.

Step 1: Find Critical Points

To find the critical points of the function, we need to set the derivative equal to zero or find where it is undefined. The derivative is given by \[ f'(x) = x^{-1/2}(x-3). \]Let's first set the derivative equal to zero:\[ x^{-1/2}(x-3) = 0. \]This implies:\[ (x-3) = 0 \Rightarrow x = 3. \]Next, identify where the derivative is undefined. Since \( x^{-1/2} \) is undefined for \( x = 0 \), this is another critical point due to the zero factor. Thus, the critical points are at \( x = 0 \) and \( x = 3 \).

Step 2: Test Intervals for Increasing/Decreasing

To determine where \( f \) is increasing or decreasing, examine the sign of \( f'(x) \) in intervals around the critical points. The critical points divide the x-axis into three intervals: \((-\infty, 0)\), \((0, 3)\), \((3, \infty)\).- For \( x \in (-\infty, 0) \), choose a test point (e.g., \( x = -1 \)). Calculating \( f'(-1) \), we findthe product is positive and undefined due to the root term, making this section not possible for analysis, as \( x^{-1/2} \) requires positive values.- For \( x \in (0, 3) \), choose \( x = 1 \):\[ f'(1) = 1^{-1/2}(1-3) = -2. \]Since the result is negative, \( f \) is decreasing on \((0, 3)\).- For \( x \in (3, \infty) \), choose \( x = 4 \):\[ f'(4) = 4^{-1/2}(4-3) = \frac{1}{2}. \]Since the result is positive, \( f \) is increasing on \((3, \infty)\).

Step 3: Determine Local Extremes

Identify any local maxima or minima using the First Derivative Test. We have sulutions for critical points:- At \( x = 0 \), since we cannot approach from negative x-values, testing was not suitable.- At \( x = 3 \), \( f'(x) \) changes from negative to positive as \( x \) crosses \( 3 \). Therefore, this is a local minimum, as function transitions from decreasing to increasing.

Key Concepts

Critical Points

Critical points of a function are where its derivative equals zero or is undefined. These points are crucial because they can indicate potential maxima, minima, or inflection points in a function. For our exercise, the derivative of the function is given by:\[ f'(x) = x^{-1/2}(x-3). \]To find the critical points, we set the derivative equal to zero:\[ x^{-1/2}(x-3) = 0. \]This equation implies two solutions when solving:- \((x-3) = 0\) leads to \(x = 3\)- \(x^{-1/2}\) is undefined at \(x = 0\)Thus, the critical points are at \(x = 3\) and an undefined point \(x = 0\). These points are where we concentrate our further analysis of the function.

Increasing Decreasing Functions

To determine where a function increases or decreases, examine the intervals around the critical points. By analyzing the sign of the derivative, we can gauge whether the function is climbing or descending:- For \(x \in (-\infty, 0)\), no valid testing due to \(x^{-1/2}\) being undefined for negative \(x\).- In \((0, 3)\), testing with \(x = 1\) gives: * \[ f'(1) = 1^{-1/2}(1-3) = -2 \] * Here, the derivative is negative, indicating a decreasing function.- For \(x \in (3, \infty)\), test with \(x = 4\): * \[ f'(4) = 4^{-1/2}(4-3) = \frac{1}{2}. \] * The positive derivative here indicates that the function is increasing.By understanding these changes, we can visualize how the function behaves across these intervals.

First Derivative Test

The First Derivative Test helps identify the nature of critical points. It involves analyzing the sign changes of the derivative around critical points:- At \(x = 0\), analysis isn't possible since negative \(x\)-values leave the derivative undefined.- At \(x = 3\), the derivative shifts from negative to positive, thus indicating a local minimum.This test, with simple sign analysis of the derivative around critical points, provides insights into where functions have peaks or troughs.

Local Maximum and Minimum

Local maxima and minima are the peaks and valleys of a function within a certain region. To identify them using critical points:- A local minimum is found at \(x = 3\) since \(f'(x)\) shifts from negative to positive.- Meanwhile, no local maxima were identified in this exercise as the derivative does not shift to negative at any critical points found.Understanding the derivative's behavior helps us pinpoint these extremes, showing how a function reaches its highs and lows locally.

Derivative Undefined

A function's derivative may be undefined at certain points, indicating possible critical points:- For \(f'(x) = x^{-1/2}(x-3)\), the expression \(x^{-1/2}\) is undefined for \(x = 0\).This is considered a critical point, but further analysis is constrained as the derivative doesn't exist in such cases for negative \(x\).Recognizing where the derivative is undefined helps identify unique points, though typical methods for finding function increases or decreases may not apply.