Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ s(t)=t^{2 / 5} ; I=[-1,32] $$

Step-by-Step Solution

Answer

Maximum value is 4 at \( t = 32 \) and minimum value is 0 at \( t = 0 \).

Step 1: Find the Derivative

To find the critical points of the function \( s(t) = t^{2/5} \), we first need to determine its derivative. Applying the power rule \( \frac{d}{dt}[t^n] = nt^{n-1} \), the derivative \( s'(t) \) is given by: \[ s'(t) = \frac{2}{5}t^{-3/5} = \frac{2}{5} \cdot \frac{1}{t^{3/5}}. \]

Step 2: Identify Critical Points

The critical points occur where the derivative is zero or undefined. Since \( \frac{2}{5} \/ t^{3/5} \) is undefined at \( t = 0 \), and never equals zero, \( t = 0 \) is a critical point.

Step 3: Evaluate the Endpoints and Critical Points

Evaluate \( s(t) \) at the endpoints \( t = -1 \) and \( t = 32 \), as well as the critical point \( t = 0 \):- \( s(-1) = (-1)^{2/5} = 1 \)- \( s(0) = (0)^{2/5} = 0 \) - \( s(32) = (32)^{2/5} = 4 \)

Step 4: Determine Maximum and Minimum Values

Compare the values obtained in Step 3:- \( s(-1) = 1 \)- \( s(0) = 0 \) - \( s(32) = 4 \) The maximum value is \( 4 \) at \( t = 32 \) and the minimum value is \( 0 \) at \( t = 0 \).

Key Concepts

Derivative in Calculus

In calculus, the derivative of a function helps us understand how that function changes—or, in other words, its rate of change. The derivative is a fundamental concept used to determine the slope of the tangent line to a function at any given point. For any function expressed as \( f(t) \), its derivative is denoted as \( f'(t) \).

To find the derivative of a function like \( s(t) = t^{2/5} \), we employ differentiation rules. One of the most widely used rules is the **Power Rule in Derivatives**, which simplifies the differentiation process for power functions. This rule states that the derivative of \( t^n \) is \( n \cdot t^{n-1} \). When we apply this rule to our example, we find that:

• \( s'(t) = \frac{2}{5}t^{\frac{-3}{5}} \), resulting from \( \frac{d}{dt}[t^{2/5}] = \frac{2}{5}\cdot t^{-3/5} \).

Understanding how to differentiate a function is the first step towards finding critical points, which are pivotal in determining a function's behavior.

Maximum and Minimum Values

Maximum and Minimum values are key components in the study of a function's behavior. These values tell us the highest and lowest points that a function can reach on a particular interval. In mathematical terms, they are often referred to as extrema.

To determine these values, we examine the function's behavior at both critical points and endpoints. **Critical points** occur at values where the derivative is zero or undefined. For example, in our case, the derivative \( s'(t) = \frac{2}{5}t^{-3/5} \) is undefined at \( t = 0 \), indicating a critical point.

After finding the critical points, we evaluate the original function at these points and the endpoints of the interval to identify which are the maximum and minimum values.

• We obtained \( s(-1) = 1 \), \( s(0) = 0 \), and \( s(32) = 4 \).

From these evaluations, it is clear that 4 is the maximum value at \( t = 32 \) and 0 is the minimum value at \( t = 0 \). Determining these values is essential for understanding the overall trend of the function's graph.

Endpoints Evaluation

When analyzing a function over a specific interval, it is crucial to consider the endpoints of that interval. Endpoints are the starting and ending values of the interval, and evaluating them provides comprehensive insights into the function's extrema. This evaluation is particularly important because extreme values can occur at these endpoints.

In the function \( s(t) = t^{2/5} \), we evaluated the endpoints \( t = -1 \) and \( t = 32 \). This calculation allows us to compare these values with those of the critical points.

• At \( t = -1 \), \( s(-1) = 1 \).
• At \( t = 32 \), \( s(32) = 4 \).

These evaluations, along with those at critical points, assist us in determining the maximum and minimum values of the function over the defined interval. Skipping this step might result in missing key information about the function's overall behavior.

Power Rule in Derivatives

The **Power Rule** is one of the fundamental tools in calculus for differentiating functions that are polynomials or power functions. It simplifies the process of finding the derivative by providing a straightforward formula. For any function expressed as \( f(t) = t^n \), the derivative is found using the power rule:

• \( f'(t) = n \cdot t^{n-1} \)

In our scenario with \( s(t) = t^{2/5} \), applying the power rule gives us the derivative:

• \( s'(t) = \frac{2}{5} \cdot t^{-3/5} \)

This rule is particularly useful because it is easy to apply, even when dealing with fractional exponents. By making differentiation more approachable, the power rule allows us to quickly find critical points, aiding in further analysis of maximum and minimum values. Understanding this rule thoroughly is key to mastering derivative calculations in calculus.