Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The function \(f(x)=\tan x\) has an inverse function when its domain is restricted to \((-\pi / 2, \pi / 2)\) (a) Graph \(y=\tan ^{-1}(x)\). Where is the derivative positive? Negative? (b) Is \(\tan ^{-1}(x)\) an even function, an odd function, or neither? (c) Is the derivative of \(\tan ^{-1}(x)\) even, odd, or neither?

Step-by-Step Solution

Answer

The derivative of \( \tan^{-1}(x) \) is always positive. The function \( \tan^{-1}(x) \) is odd while its derivative is even.

Step 1: Graph the inverse function \( \tan^{-1}(x) \)

The graph of \( y = \tan^{-1}(x) \) is obtained in a similar manner as \( y = \tan(x) \) with the domain and range flipped. That is, for \( y = \tan^{-1}(x) \), the domain is all real numbers and the range is \( (- \pi / 2, \pi / 2) \). Like \( y = \tan(x) \), it also has a curve that looks like the letter 'S'.

Step 2: Determine where the derivative is positive and negative

To derive \( y = \tan^{-1}(x) \) rather, \( y' = 1/(1+x^2) \). This derivative is always positive for all \( x \) because the numerator is always positive and the denominator as sum of squares is also always positive.

Step 3: Determine if \( \tan^{-1}(x) \) is even, odd, or neither

An even function is one where \( f(x) = f(-x) \). An odd function is one where \( -f(x) = f(-x) \). Thus, for \( \tan^{-1}(x) \), it can be observed that \( \tan^{-1}(-x) = - \tan^{-1}(x) \), suggesting that \( \tan^{-1}(x) \) is an odd function.

Step 4: Determine if the derivative of \( \tan^{-1}(x) \) is even, odd, or neither

If the derivative of \( \tan^{-1}(x) \), that is, \( 1/(1+x^2) \), \( f'(x) = f'(-x) \), the function is even. Thus, the derivative of \( \tan^{-1}(x) \) is an even function.

Key Concepts

Trigonometric Functions

Trigonometric functions are mathematical functions of an angle. They are vital in calculations involving periodic phenomena, such as waves. One such function, the tangent function denoted as \( \tan(x) \), has special properties and applications. It is defined as the ratio of the sine and cosine functions: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \).
The function is undefined where \( \cos(x) = 0 \), owing to the division by zero, so it has vertical asymptotes at these points. The tangent function has a periodicity of \( \pi \), meaning its graph repeats every \( \pi \) units.
When we talk about the inverse of the tangent function, \( \tan^{-1}(x) \) or arctan, it is crucial to consider its domain and range. The inverse function has a restricted domain of \((-\pi/2, \pi/2)\), allowing it to pass the horizontal line test, making it a one-to-one function.

• Domain of \( \tan^{-1}(x) \): All real numbers.
• Range of \( \tan^{-1}(x) \): \((-\pi/2, \pi/2)\).

Its graph typically resembles a smooth 'S' shape, reflecting around the origin when plotted on a Cartesian plane.

Derivatives

Derivatives represent the rate at which a function is changing at any given point, often interpreted as the slope of the function’s graph. The derivative is a fundamental concept in calculus that provides insights into the behavior of functions.
For the inverse tangent function \( y = \tan^{-1}(x) \), the derivative is \( y' = \frac{1}{1+x^2} \). This derivative describes the slope of \( \tan^{-1}(x) \) for any value of \( x \), and it plays an essential role in determining the behavior of this function throughout its domain.

• The expression \( \frac{1}{1+x^2} \) indicates that the derivative is always positive because both the numerator is positive (a constant \( 1 \)) and the denominator is the sum of squares, which is always positive.
• This positive derivative confirms that \( \tan^{-1}(x) \) is increasing for all real \( x \).

Understanding where a derivative is positive or negative can reveal where a function is increasing or decreasing, which is a key aspect of graph analysis.

Odd and Even Functions

Odd and even functions have specific symmetrical properties that make them unique. These symmetries relate to the powers of \( x \) in their polynomial expansions.
For a function to be even, such as \( f(x) = f(-x) \), it must be symmetric about the y-axis. Conversely, odd functions satisfy the condition \( f(-x) = -f(x) \), resulting in symmetry about the origin.
The function \( \tan^{-1}(x) \) is classified as an odd function because:

• It satisfies the equation \( \tan^{-1}(-x) = -\tan^{-1}(x) \), indicating origin symmetry.

This property is visually represented on its graph, where the curve appears the same when flipped across the origin.

Now, regarding the derivative of \( \tan^{-1}(x) \), \( \frac{1}{1+x^2} \), this function is even. To determine evenness for the derivative, we test \( f'(x) = f'(-x) \), and indeed find \( \frac{1}{1+x^2} = \frac{1}{1+(-x)^2} \). Therefore, the derivative exhibits symmetry about the y-axis, completing the picture of how the inverse tangent function behaves in terms of odd and even symmetry.