Problem 21 - Calculus: Early Transcendentals

Question

Find the absolute extrema for \(f(x)=x-\tan ^{-1}(2 x)\) on [0,2] .

Step-by-Step Solution

Answer

The absolute maximum is approximately 0.675 at \(x = 2\), and the minimum is approximately -0.285 at \(x = \frac{1}{2}\).

Step 1: Understand the Function and Domain

The function given is \(f(x) = x - \tan^{-1}(2x)\) and we need to find its absolute extrema on the interval [0,2]. The extrema refer to the highest and lowest values the function achieves in this interval.

Step 2: Calculate the Derivative

To find the critical points, we first differentiate \(f(x)\). The derivative \(f'(x)\) is given by:\[\frac{d}{dx}\left(x - \tan^{-1}(2x)\right) = 1 - \frac{2}{1 + (2x)^2}.\]Simplify to:\[f'(x) = 1 - \frac{2}{1 + 4x^2}.\]

Step 3: Find Critical Points

Set the derivative \(f'(x)\) to zero and solve:\[1 - \frac{2}{1 + 4x^2} = 0.\]This simplifies to:\[1 = \frac{2}{1 + 4x^2}.\]Solving it, we find:\[1 + 4x^2 = 2\]\[4x^2 = 1\]\[x^2 = \frac{1}{4}\]\[x = \frac{1}{2}.\]Thus, the critical point is \(x = \frac{1}{2}\).

Step 4: Evaluate the Function at Critical and Endpoint Values

Evaluate \(f(x)\) at the critical point and the interval endpoints.- \(f(0) = 0 - \tan^{-1}(0) = 0\)- \(f\left(\frac{1}{2}\right) = \frac{1}{2} - \tan^{-1}(1) = \frac{1}{2} - \frac{\pi}{4}\)- \(f(2) = 2 - \tan^{-1}(4)\)

Step 5: Compare Function Values to Find Extrema

Calculate the approximate values:- \(f(0) = 0\)- To calculate \(f\left(\frac{1}{2}\right)\), note \(\tan^{-1}(1) = \frac{\pi}{4} \approx 0.785\), so \(f\left(\frac{1}{2}\right) \approx \frac{1}{2} - 0.785 \approx -0.285\)- \(\tan^{-1}(4)\) is approximately 1.325 (from a calculator), so \(f(2) = 2 - 1.325 \approx 0.675\).Thus, the highest value is at \(f(2)\) and the lowest is at \(f\left(\frac{1}{2}\right)\).

Key Concepts

Absolute Extrema

When solving for absolute extrema, you want to find the highest and lowest values a function takes on a specific interval. Here, we're looking at the function \(f(x) = x - \tan^{-1}(2x)\) over the interval \([0,2]\). These extrema occur either at critical points inside the interval or at the endpoints.

• To determine this, evaluate the function at both critical points and the endpoints of the interval.
• This will give you all possible candidates for absolute maximum and minimum values.

For \(f(x)\), the endpoints are \(x = 0\) and \(x = 2\), and the critical point was found at \(x = \frac{1}{2}\). Evaluating \(f\) at these points helps us determine which of these is the function's highest (maximum) and lowest (minimum) value on that interval.

Critical Points

Critical points of a function occur where the derivative is zero or undefined. They are important because they are potential locations of local maxima, minima, or turning points.
For \(f(x) = x - \tan^{-1}(2x)\), we calculate the derivative \(f'(x) = 1 - \frac{2}{1 + 4x^2}\) and set it equal to zero to find critical points:

• \(1 - \frac{2}{1 + 4x^2} = 0\)

Solving this equation, we find that the critical point is at \(x = \frac{1}{2}\).
This point is significant because it could potentially be where the function changes direction, leading to a local or absolute minimum or maximum. Evaluating the function at this point helps us determine its behavior in the context of the interval \([0,2]\).

Derivative Calculations

Calculating derivatives is a fundamental part of analyzing the behavior of functions. In this exercise, we needed the derivative \(f'(x)\) of \(f(x) = x - \tan^{-1}(2x)\) to find critical points.
The process involves applying derivative rules:

• The derivative of \(x\) is \(1\).
• For \(\tan^{-1}(2x)\), use the chain rule. The derivative of \(\tan^{-1}(u)\) is \(\frac{1}{1+u^2}\) where \(u = 2x\).
  • Then, multiply by the derivative of \(u\), which is \(2\).

Putting it all together, \(f'(x) = 1 - \frac{2}{1 + 4x^2}\). This derivative helps locate critical points and analyze the slope and behavior of \(f(x)\) throughout the interval. Derivatives provide crucial insights into how a function is increasing or decreasing.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\tan^{-1}(x)\), are used to reverse the process of trigonometric functions. They are crucial when working with trigonometric identities or solving equations involving angles.
In this problem, the inverse arctan function, \(\tan^{-1}(2x)\), is part of the original function \(f(x) = x - \tan^{-1}(2x)\).

• \(\tan^{-1}(x)\) gives the angle whose tangent is \(x\).
• This is important for transforming the trigonometric relationship back to an angle measure.

When analyzing its role within the function, inverse trigonometric functions affect how \(f(x)\) behaves since they introduce non-linear variation as the input \(2x\) changes over the interval \([0,2]\). Recognizing the contribution of \(\tan^{-1}(2x)\) aids in fully understanding the function's structure and the behavior we study in calculus.