Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ f(s)=|3 s-2| ; I=[-1,4] $$

Step-by-Step Solution

Answer

The minimum value is 0 at \( s = \frac{2}{3} \) and the maximum value is 10 at \( s = 4 \).

Step 1: Understand the function

The function given is \( f(s) = |3s - 2| \). This is an absolute value function, which means its graph will be V-shaped. We need to find where the expression inside the absolute value changes from positive to negative, or vice versa.

Step 2: Find the critical points

To find where \( f(s) \) changes its direction, solve \( 3s - 2 = 0 \). Solving this equation gives \( s = \frac{2}{3} \). This is the critical point where the function changes direction.

Step 3: Evaluate the endpoints

Evaluate the function at the endpoints of the interval \([-1, 4]\). Calculate \( f(-1) \) and \( f(4) \): 1. \( f(-1) = |3(-1) - 2| = | -3 - 2| = 5 \).2. \( f(4) = |3(4) - 2| = |12 - 2| = 10 \).

Step 4: Evaluate the critical point

Evaluate the function at the critical point \( s = \frac{2}{3} \):\( f(\frac{2}{3}) = |3(\frac{2}{3}) - 2| = |2 - 2| = 0 \).

Step 5: Determine the maximum and minimum values

From the evaluations: \( f(-1) = 5 \), \( f(\frac{2}{3}) = 0 \), and \( f(4) = 10 \). The minimum value is 0 (at \( s = \frac{2}{3} \)) and the maximum value is 10 (at \( s = 4 \)).

Key Concepts

Absolute Value Functions

Absolute value functions are mathematical expressions that involve the absolute value operation, denoted by vertical bars \(| \cdot |\). The absolute value of a number represents its distance from zero on the number line. For any real number \(x\), the absolute value \(|x|\) is defined as:

• \(x\) if \(x \geq 0\)
• \(-x\) if \(x < 0\)

In calculus, an absolute value function often appears as part of an expression, such as \(f(s) = |3s - 2|\). These functions create V-shaped graphs, as the absolute value causes any negative input to become non-negative. This results in a corner or 'cusp' at the point of reflection, where the direction of the slope changes.

Maximum and Minimum Values

The maximum and minimum values of a function provide critical insights into its behavior within a given interval. For a function \(f(s) = |3s - 2|\), identifying these values helps understand the height and depth of the graph on the interval \([-1, 4]\). Calculating these values involves evaluating the function at critical points and endpoints. Critical points occur where the inside expression, \(3s - 2\), switches signs. To find these, we solve \(3s - 2 = 0\), yielding \(s = \frac{2}{3}\) as the critical point. Consequently, you compare function values at the endpoints \(s = -1\) and \(s = 4\) together with the critical point to determine the highest and lowest values. For this function, values are \(f(-1) = 5\), \(f(\frac{2}{3}) = 0\), and \(f(4) = 10\), which reveals 0 as the minimum and 10 as the maximum.

Evaluating at Endpoints

Evaluating a function at its endpoints is crucial to finding absolute maximum and minimum values over a specific interval. Endpoints represent the boundary values where the function's interval begins and ends. For \(f(s) = |3s - 2|\) on the interval \([-1, 4]\), the endpoints are \(s = -1\) and \(s = 4\). Evaluating at these points involves substituting them into the function:

• For \(s = -1\), calculate \(f(-1) = |3(-1) - 2| = 5\).
• For \(s = 4\), calculate \(f(4) = |3(4) - 2| = 10\).

This process helps ensure no potential maximum or minimum value is overlooked, especially if it occurs at the boundary rather than at a critical point inside the interval.

V-shaped Graphs

V-shaped graphs are characteristic of absolute value functions, like \(f(s) = |3s - 2|\). These graphs feature a distinct vertex, the lowest or highest point of the V, where the slope changes direction. The graph is made of two linear segments:

• One segment has a positive slope extending from the vertex forward on the graph.
• The other has a negative slope extending backward.

In the case of \(f(s) = |3s - 2|\), the vertex occurs at \(s = \frac{2}{3}\), where the expression \(3s - 2\) equals zero. V-shaped graphs help visually identify critical points and anticipate where maximum and minimum values might occur based on the graph's rise and fall around the vertex.