Problem 33 - Thomas Calculus

Question

In Exercises \(31-34,\) find the function's absolute maximum and minimum values and say where they are assumed. $$ g(\theta)=\theta^{3 / 5}, \quad-32 \leq \theta \leq 1 $$

Step-by-Step Solution

Answer

Max: 1 at \(\theta = 1\); Min: -8 at \(\theta = -32\).

Step 1: Find Critical Points

First, we need to find the critical points of the function \( g(\theta) = \theta^{3/5} \). A critical point occurs where the derivative is zero or undefined. Find \( g'(\theta) \). Use the power rule to differentiate: \[ g'(\theta) = \frac{3}{5} \theta^{-2/5} = \frac{3}{5\theta^{2/5}} \]The derivative is undefined at \( \theta = 0 \) since we cannot have a zero in the denominator. This is our critical point.

Step 2: Evaluate Function at Critical Points

We evaluate the function \( g(\theta) \) at the critical point found in Step 1:\[ g(0) = 0^{3/5} = 0 \]

Step 3: Evaluate Function at Endpoints

Next, evaluate \( g(\theta) \) at the endpoints of the interval, \(-32\) and \(1\):\[ g(-32) = (-32)^{3/5} = -8 \] (since \((-32)^{3/5} = (-(32^{1/5}))^3 = -2^3 = -8)\)\[ g(1) = 1^{3/5} = 1 \]

Step 4: Compare Values

Compare the values obtained in Steps 2 and 3 to determine the absolute maximum and minimum values:- \( g(-32) = -8 \)- \( g(0) = 0 \)- \( g(1) = 1 \)The absolute maximum value is 1 at \( \theta = 1 \), and the absolute minimum value is -8 at \( \theta = -32 \).

Key Concepts

Critical Points

Critical points are crucial in the optimization of functions as they potentially indicate where a function can change from increasing to decreasing, or vice versa. To find critical points, we focus on the derivative of the function, denoted as \( g'(\theta) \). Critical points occur where this derivative is either zero or undefined.

For the given function \( g(\theta) = \theta^{3/5} \), we computed the derivative:

• Using the power rule, this derivative is \( g'(\theta) = \frac{3}{5} \theta^{-2/5} = \frac{3}{5\theta^{2/5}} \).

Here, observe that the derivative becomes undefined at \( \theta = 0 \) since division by zero is impossible. Thus, \( \theta = 0 \) is identified as a critical point. These points are significant, as they are candidates for local extreme values and can help determine absolute extremum values within a given interval, in this case, \(-32 \leq \theta \leq 1\).

Absolute Maximum and Minimum

To determine the absolute maximum and minimum values of a function, we analyze the function at critical points and the endpoints of a closed interval. This comprehensive evaluation ensures that no maximum or minimum values are missed.

For the function \( g(\theta) = \theta^{3/5} \), we checked:

• The critical point found was at \( \theta = 0 \) giving us a function value \( g(0) = 0 \).
• The endpoints of the interval \( \theta = -32 \) and \( \theta = 1 \) gave us \( g(-32) = -8 \) and \( g(1) = 1 \), respectively.

After computing these values, compare them to find the absolute extremum:

• The maximum value is \( 1 \) at \( \theta = 1 \).
• The minimum value is \(-8\) at \( \theta = -32 \).

Thus, within the given interval, \( g(\theta) \) reaches its absolute highest and lowest points.

Derivative

The derivative of a function provides valuable information about its rate of change at any given point. It is a fundamental concept in calculus used widely for analyzing and optimizing functions.

In our exercise, to find the derivative of \( g(\theta) = \theta^{3/5} \), we applied the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). This resulted in:

• \( g'(\theta) = \frac{3}{5} \theta^{-2/5} = \frac{3}{5\theta^{2/5}} \)

A key aspect here is recognizing where the derivative does not exist, such as at \( \theta = 0 \), which becomes essential for finding critical points for optimization. This makes derivatives a powerful tool for identifying changes in the behavior of functions, particularly regarding optimization tasks and finding extreme values within an interval.