Problem 31 - Thomas Calculus

Question

In Exercises \(31-34,\) find the function's absolute maximum and minimum values and say where they are assumed. $$ f(x)=x^{4 / 3}, \quad-1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

The absolute minimum is 0 at x=0; the absolute maximum is 16 at x=8.

Step 1: Identify the Function and Its Domain

We are given the function \( f(x) = x^{4/3} \) with the domain \(-1 \leq x \leq 8\). Our task is to find the absolute maximum and minimum values of this function over the closed interval \(-1, 8\).

Step 2: Check Endpoints of the Interval

Evaluate the function at the endpoints of the domain. First, calculate \( f(-1) = (-1)^{4/3} = 1 \). Then, evaluate \( f(8) = 8^{4/3} = 16 \).

Step 3: Find Critical Points in the Interval

To find critical points, calculate the derivative of \( f(x) \). The derivative is given by \( f'(x) = \frac{4}{3}x^{1/3} \). Set \( f'(x) = 0 \) to find critical points: \( \frac{4}{3}x^{1/3} = 0 \) implies \( x = 0 \). Thus, \( x = 0 \) is a critical point. Check if it is within the interval: it is, since \(-1 \leq 0 \leq 8\).

Step 4: Evaluate the Function at the Critical Point

Evaluate the function at the critical point \( x = 0 \): \( f(0) = 0^{4/3} = 0 \).

Step 5: Compare Values to Determine Extremes

We now compare the function values at the endpoints and the critical point: \( f(-1) = 1 \), \( f(0) = 0 \), and \( f(8) = 16 \). The absolute minimum value is \( 0 \) at \( x = 0 \), and the absolute maximum value is \( 16 \) at \( x = 8 \).

Key Concepts

Critical Points

Understanding critical points is essential when searching for absolute extrema in a function. A critical point occurs where the derivative of the function equals zero or is undefined. These points are significant because they might indicate a local maximum, local minimum, or saddle point, which could potentially be where the absolute extrema are located in a given domain.

To find the critical points of a function, you need to compute its derivative and solve for the values that make the derivative zero or undefined. In our example, with the function \( f(x) = x^{4/3} \), the derivative \( f'(x) = \frac{4}{3}x^{1/3} \) equals zero when \( x = 0 \). This value represents a critical point. It is crucial to ensure that the critical point lies within the closed interval being considered—in this case, \(-1\) to \(8\).

By evaluating the function at these critical points, along with the endpoints of the interval, we can determine where the function reaches its absolute maximum and minimum values.

Derivative

The derivative of a function gives us the rate of change of the function at any point along its curve. Calculating the derivative is a vital step in finding absolute extrema because it helps identify critical points, where potential extremes occur.

For the function \( f(x) = x^{4/3} \), we calculate the derivative to be \( f'(x) = \frac{4}{3}x^{1/3} \). This derivative tells us how the function's value changes with small changes in \( x \). A derivative equal to zero can indicate a turning point or critical point in the graph of the function.

When you set the derivative equal to zero, you're looking for flat spots on the curve where the function might change direction. However, not all critical points are points of absolute extrema. This is why further evaluation of the function values is necessary.

Function Evaluation

Function evaluation involves substituting specific values into the function to find corresponding outputs. This method is particularly useful when determining the absolute maxima and minima within a specific interval.

In solving for the absolute extrema of \( f(x) = x^{4/3} \) on the interval \(-1 \leq x \leq 8\), we evaluate:

• The function at the endpoints: \( f(-1) = 1 \) and \( f(8) = 16 \)
• The function at the critical point: \( f(0) = 0 \)

By comparing these values, we determine that the smallest value (or absolute minimum) is \( 0 \), occurring at \( x = 0 \), and the largest value (or absolute maximum) is \( 16 \), occurring at \( x = 8 \).

Function evaluation is a straightforward but crucial step in the process of finding absolute extrema, as it helps pinpoint the exact values of interest on the interval.

Endpoints

Endpoints are the boundaries of the interval considered for finding absolute extrema. Evaluating the function at these points is a must-do step because absolute maximum or minimum values can occur here, rather than at critical points.

When solving an optimization problem, such as finding the extrema of \( f(x) = x^{4/3} \) from \(-1 \leq x \leq 8\), checking the function values at the endpoints \(-1\) and \(8\) is necessary.

By calculating \( f(-1) = 1 \) and \( f(8) = 16 \), we established that the absolute maximum within the interval occurs at \( x = 8 \), not at any of the critical points found within the same domain. On occasion, the extrema may align with these boundary evaluations, demonstrating that endpoints play an equally significant role in solving for absolute extrema.