Question

In Exercises, find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. $$ f(x)=3 x^{2 / 3}-2 x, \quad[-1,2] $$

Short answer

The absolute extrema of the function \(f(x) = 3x^{2/3} - 2x\) on the interval \([-1, 2]\) is the maximum value of 5 at \(x = -1\) and the minimum value of 0 at \(x = 2\).

Step-by-step solution

Find the derivative of the function

Differentiate the given function \(f(x) = 3x^{2/3} - 2x\). Using the power rule, \(f'(x) = 2x^{-1/3} - 2\).

Find critical points

Set the derivative equal to zero and solve for x to find the critical points. \(2x^{-1/3} - 2 = 0\) implies \(x^{-1/3} = 1\), which gives \(x = 1\). Since \(x = 1\) is in the interval \([-1, 2]\), it is a critical point.

Evaluate the function at end points and the critical point

Evaluate the function at the end points of the interval, -1 and 2, and at the critical point 1. \(f(-1) = 3(-1)^{2/3} - 2(-1) = 3+2 = 5,\) \(f(1) = 3(1)^{2/3} - 2(1) = 3-2 = 1,\) \(f(2) = 3(2)^{2/3} - 2(2) = 4-4 = 0.\)

Identify the absolute extrema

By comparing the function values at -1, 1, and 2, clearly, the absolute maximum of the function on the interval \([-1, 2]\) is 5 at \(x = -1\) and the absolute minimum is 0 at \(x = 2\).

Key concepts

Critical Points

Critical points are essential in calculus to help locate where a function reaches its potential maximum or minimum values within a given domain.

These points occur when the first derivative of a function equals zero or is undefined. For the function evaluated in the exercise, the critical point was found by setting the derivative, \(f'(x) = 2x^{-1/3} - 2\), to zero. Solving \(x^{-1/3} = 1\) resulted in the critical point \(x = 1\).

Within the closed interval \([-1, 2]\), the critical point at \(x = 1\) was used along with the endpoints to determine where the function reached its absolute extrema.

• Setting the derivative to zero helps to find critical points.
• Evaluate points where the derivative is undefined.
• Ensure critical points fall within the given interval.

Derivative

The concept of a derivative is central to calculus and provides information about the rate of change of a function.

In the given exercise, finding the derivative of the function \(f(x) = 3x^{2/3} - 2x\) was critical. The power rule, a basic differentiation technique, was applied. By differentiating, the function obtained was \(f'(x) = 2x^{-1/3} - 2\).

The derivative helps in identifying the behavior of the function, such as determining where the slope of the function graph changes or becomes flat. This directly leads to locating critical points, which are vital in analyzing the function's extrema.

• The derivative indicates where a function increases or decreases.
• Applying the power rule simplifies finding derivatives for polynomial terms.
• The derivative must be set to zero to find potential extrema.

Closed Interval

The closed interval represents all the numbers between two endpoints, including the endpoints themselves.

For the exercise, the closed interval \([-1, 2]\) included both \(-1\) and \(2\). This means when searching for absolute extrema, both these endpoints had to be considered along with any critical points within the interval.

Closed intervals are crucial in calculus when determining where a function is evaluated or constrained. One graphical implication is that both ends of the interval are points on the graph that can potentially be maxima or minima.

• The endpoints of a closed interval are included in evaluations.
• Closed intervals indicate the domain over which you analyze a function.
• Both endpoints and internal critical points must be checked for extrema.

Function Evaluation

Function evaluation is the process of determining the outputs of a function for specific input values.

In the exercise, the function \(f(x)\) was evaluated at the critical point and the endpoints of the interval. Specifically, it was calculated at \(x = -1, 1, \) and \(2\).

This step of evaluating functions at specific points is key in finding absolute extrema. By comparing the function values at each of these points, it was determined which points were absolute maxima or minima.

This comparison allowed determining that the maximum function value was 5 at \(x = -1\), and the minimum was 0 at \(x = 2\).

• Evaluate the function at critical points and endpoints for extremum analysis.
• Compare the calculated function values to identify maximum and minimum values.
• Function evaluation is crucial when analyzing the behavior and outcomes of a function.