Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, 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

Critical point: \(s = \frac{2}{3}\); Min value: 0 at \(s = \frac{2}{3}\); Max value: 10 at \(s = 4\).

Step 1: Understand Absolute Function

The function given is of the form \( f(s) = |3s - 2| \). An absolute value function can make different linear pieces depending on where the argument is zero.

Step 2: Identify Critical Point for Absolute Value

First, set the inside of the absolute value equal to zero to find any critical point: \[3s - 2 = 0\]Solving this equation gives:\[s = \frac{2}{3}\].This is a critical point where the function may change slope.

Step 3: Evaluate Endpoints and Critical Point

Evaluate \(f(s)\) at the critical point \(s = \frac{2}{3}\) and the endpoints \(s = -1\) and \(s = 4\):- \( f(-1) = |3(-1) - 2| = |-5| = 5\)- \( f\left(\frac{2}{3}\right) = |3\left(\frac{2}{3}\right) - 2| = |2 - 2| = 0\)- \( f(4) = |3(4) - 2| = |12 - 2| = 10\).

Step 4: Determine Maximum and Minimum Values

Based on the evaluations:- The minimum value is \(0\) at \(s = \frac{2}{3}\).- The maximum value is \(10\) at \(s = 4\).

Key Concepts

Understanding the Absolute Value Function

The absolute value function, represented as \( f(s) = |3s - 2| \), involves an operation that affects how values appear. The absolute value denotes the distance of a number from zero on a number line, without consideration of direction.
Mathematically, \( |x| \) can be broken down as follows:

• If \( x \geq 0 \), then \( |x| = x \).
• If \( x < 0 \), then \( |x| = -x \).

In our function \( f(s) = |3s - 2| \), we must first determine the values of \( s \) for which \( 3s - 2 = 0 \). This schism will guide us into breaking down the function into linear segments or pieces.

Identifying Critical Points in Calculus

Critical points are essential because they indicate where the function's behavior changes, particularly in terms of slope or direction. For the absolute value function \( f(s) = |3s - 2| \), identifying critical points means finding \( s \) where the function shifts its linear segment.
The calculation starts with setting the inside of the absolute function to zero:\[ 3s - 2 = 0 \]Solving gives \( s = \frac{2}{3} \), making \( s = \frac{2}{3} \) a critical point.
At this point, \( s \)'s behavior transitions, representing a possible point of minimum or maximum in the context of the function on a given interval.

Finding Maximum and Minimum Values

Maximum and minimum values in calculus are locations where the function peaks and dips within a defined section, often between endpoints or at critical points. To find these values for \( f(s) = |3s - 2| \) on \( s \in [-1, 4] \), check key spots:

• Endpoints \( s = -1 \) and \( s = 4 \)
• Critical point \( s = \frac{2}{3} \)

Evaluating yields:

• \( f(-1) = 5 \)
• \( f(\frac{2}{3}) = 0 \)
• \( f(4) = 10 \)

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

Evaluating Functions at Critical Points

Evaluating functions at critical points reveals important insights about their behavior at these key points. Given the function \( f(s) = |3s - 2| \) with a critical point at \( s = \frac{2}{3} \), evaluating \( f \) here provides the lowest value on the interval. This evaluation is crucial for confirming if the critical point corresponds to a minimum or maximum.
Understanding that the function reaches a zero value highlights a turnaround moment on the graph's slope.
Calculating \( f \) at critical points and comparing them with endpoints helps in identifying such points.

Exploring End Behavior in Calculus

End behavior in calculus describes how a function behaves as \( s \) approaches the boundaries of its domain or interval. For the absolute value function \( f(s) = |3s - 2| \), observing end behavior involves evaluating the function at the endpoints of the interval \( s \in [-1, 4] \).
By checking:

• \( f(-1) = 5 \)
• \( f(4) = 10 \)

We can deduce that as \( s \) moves from \(-1\) to \(4\), the function starts at \(5\), reaches a minimum at zero, then peaks again at \(10\). This analysis of starting and ending values complements your understanding of the function's range and provides a complete view of its behavior over the interval.