Problem 23 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ y=3 x^{2 / 3}-2 x,[-1,1] $$

Step-by-Step Solution

Answer

The function \(y = 3x^{2 / 3} - 2x\) has an absolute maximum at \(x = -1\) and an absolute minimum at \(x = 0\).

Step 1: Differentiate

Differentiate the function to find its derivative. \(y' = (2/3)3x^{-1/3}-2 =2x^{-1/3}-2\). This function is used to identify the critical points.

Step 2: Solve Derivative for Zero and Undefined

To find the critical points, make the derivative equal to zero and solve the equation: \(2x^{-1/3}=2\). x = 1 is a critical point. For the derivative being undefined: this requires \(x = 0\) which is also within the interval.

Step 3: Find Function Values at Critical Points and Endpoints

Evaluate the original function at the critical points and at the endpoints of the interval. This will give us four values: \(f(-1)\) = \(3(-1)^{2/3}-2(-1) = 5\), \(f(0) = 0\), \(f(1) = 1\) and \(f(1) = 1\).

Step 4: Identify the Absolute Maximum and Minimum

The absolute maximum is the largest of the four values, seen to be at \(x = -1\). The least one is the absolute minimum value, which is 0 when \(x = 0\).

Key Concepts

Critical Points

Critical points are like the roadmap to understanding where a function might have peaks or valleys, also called extrema. They occur when the first derivative of a function equals zero or becomes undefined. Essentially, you're looking for spots where the slope of the tangent to the graph is either flat or does not exist.

There are two cases for critical points:

• Derivative equals zero. This happens at points where the graph of the function has a horizontal tangent. For the function given, setting the derivative \[2x^{-1/3} - 2 = 0\] gives \(x = 1\) as a critical point.
• Derivative is undefined. Sometimes, the derivative doesn't exist for certain values. This occurs when operations like division by zero happen. In the provided exercise, \(2x^{-1/3}\) becomes undefined when \(x = 0\). Thus, \(x = 0\) is another critical point.

Understanding critical points helps locate the crucial aspects of the function, as they might indicate where the function reaches its highest or lowest points within the interval.

Derivatives

A derivative is a powerful tool in calculus that reveals a function's rate of change at any given moment. It tells us how fast or slow a function is changing its value at a particular point. In mathematical terms, for a function \(y = 3x^{2/3} - 2x\), its derivative \(y'\) is calculated to be:

• \[y' = (2/3)3x^{-1/3} - 2 = 2x^{-1/3} - 2\]

This derivative gives us information about the slope or steepness of the tangent line to the graph at any point \(x\).

By finding where the derivative equals zero or is undefined, we can investigate the behavior of the function: where it increases, decreases, or levels out.

• When the derivative is positive, the function is increasing.
• When the derivative is negative, the function is decreasing.
• When the derivative equals zero, the function might achieve a local maximum, minimum, or an inflection point where the curve changes direction.

Discovering these characteristics helps us understand not just where critical points occur, but also informs about overall trends in the function's behavior.

Absolute Maximum and Minimum

In the context of a closed interval, finding absolute maximum and minimum values of a function is crucial to understanding its entire range of behavior on a specific segment of the x-axis. Essentially, these are the highest and lowest points of the function over the given interval.

To find them, follow these steps:

• Evaluate the function at all the critical points determined previously. These are potential candidates for extrema.
• Evaluate the function at the endpoints of the intervals, since extrema can also occur at these boundaries. For instance, in the exercise, the values computed were: \[f(-1) = 5,\] \[f(0) = 0,\] \[f(1) = 1\]
• Among these, identify the largest and smallest function values. Here, the highest, \(f(-1) = 5\), is the absolute maximum, and the lowest, \(f(0) = 0\), is the absolute minimum.

Knowing where a function hits its lowest low and highest high within a specific interval is valuable for understanding its behavior fully. This insight applies to many applications, from optimizing business strategies to determining optimal parameters in various scientific fields.