Problem 50 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=x-2 \tan ^{-1} x, \quad[0,4]$$

Step-by-Step Solution

Answer

The absolute minimum is \(f(1) = 1 - \frac{\pi}{2}\); the absolute maximum is at \(f(4)\).

Step 1: Find the Derivative of the Function

To find the critical points, we first need the derivative of the function. Let's start by differentiating \(f(x) = x - 2 \tan^{-1}(x)\). Use the fact that the derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1+x^2}\). So, the derivative \(f'(x)\) is given by:\[ f'(x) = 1 - \frac{2}{1 + x^2} \]

Step 2: Solve f'(x) = 0 for Critical Points

Set the derivative \( f'(x) \) equal to zero to find critical points:\[ 1 - \frac{2}{1 + x^2} = 0 \]Re-arrange and solve for \(x\):\[ \frac{2}{1 + x^2} = 1 \]\[ 2 = 1 + x^2 \]\[ x^2 = 1 \]\[ x = 1 \]Since we are working within the interval \([0, 4]\), the critical point we consider is \(x = 1\).

Step 3: Evaluate f(x) at Critical and Endpoint Values

We need to evaluate \(f(x)\) at the critical point and at the endpoints of the interval \([0, 4]\):- \(f(0) = 0 - 2\tan^{-1}(0) = 0\)- \(f(1) = 1 - 2\tan^{-1}(1) = 1 - 2 \cdot \frac{\pi}{4} = 1 - \frac{\pi}{2}\)- \(f(4) = 4 - 2\tan^{-1}(4)\) (calculate \(\tan^{-1}(4)\) to find the exact value)These values will help us identify the absolute maximum and minimum.

Step 4: Compare the Values to Find Absolute Maximum and Minimum

Let's compare the calculated \(f(x)\) values to find the absolute maximum and minimum on the interval \([0, 4]\):- \(f(0) = 0\)- \(f(1) = 1 - \frac{\pi}{2}\)- \(f(4)\) (approximate since we need \(\tan^{-1}(4)\))From the three calculated values, the smallest value will be the absolute minimum, and the largest value will be the absolute maximum. However, let's use a calculator for the exact value at \(f(4)\) for comparison.

Key Concepts

Critical Points

In calculus, critical points are crucial for understanding a function's behavior and for solving optimization problems. A critical point occurs where the derivative of a function is zero or undefined, indicating potential points where a function may have a maximum, minimum, or neither on a given interval.
To find critical points, follow these steps:

• First, differentiate the function, obtaining its derivative.
• Next, set the derivative equal to zero to solve for the critical points.
• Ensure that any critical points found are within the interval you are exploring.

In the exercise provided, we first found the derivative of the function and solved it to determine that the critical point, within the interval [0, 4], is at \(x = 1\). This point becomes significant when evaluating the function's behavior, especially concerning absolute maximum and minimum values.
Watch out for places where the derivative is undefined as well, as these spots can also be critical points or need consideration.

Derivative

The derivative of a function provides valuable insight into its rate of change. It is a fundamental tool in calculus used to find slopes of tangent lines, and more importantly, it helps identify critical points for optimization. The process of finding a function's derivative relies on rules such as the power rule, product rule, chain rule, and others.
In this problem, we differentiated the function \(f(x) = x - 2 \tan^{-1}(x)\) using the basic rule for inverse trigonometric functions. Knowing that the derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1+x^2}\), we applied it:

• The derivative of \(x\) is 1.
• Thus, the derivative \(f'(x)\) resulted in \(1 - \frac{2}{1 + x^2}\).

This derivative was then used to find where \(f'(x) = 0\), indicating possible locations for critical points. Without taking a derivative, it would have been impossible to perform further steps like finding absolute extrema.

Absolute Maximum and Minimum Values

Absolute maximum and minimum values help determine the highest and lowest points a function can reach within a specified range, respectively. These values are crucial when making assessments, like determining the best or least outcome achievable within given constraints.
Finding these values involves:

• Identifying the derivative to locate the critical points.
• Evaluating the function at these points and at the endpoints of the interval.
• Comparing these results to figure out which values are the largest and smallest.

In our exercise, after calculating the function's values at the important points:

• At the left endpoint \(x = 0\), \(f(0) = 0\).
• At the critical point \(x = 1\), \(f(1) = 1 - \frac{\pi}{2}\).
• At the right endpoint \(x = 4\), we needed the value of \(\tan^{-1}(4)\), hence an approximate value for \(f(4)\) was used for comparison.

Comparing these calculated values determined where the absolute maximum and minimum values are located on the interval \([0, 4]\). It's important to note that extremes can occur at either critical points or at the boundaries of the interval.