Question

Let the function \(g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) be given by \(g(u)=2 \tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2} .\) Then, \(g\) is (A) even and is strictly increasing in \((0, \infty)\) (B) odd and is strictly decreasing in \((-\infty, \infty)\) (C) odd and is strictly increasing in \((-\infty, \infty)\) (D) neither even nor odd, but is strictly increasing in \((-\infty, \infty)\)

Short answer

Function g(u) is odd and strictly increasing in \((-\infty, \infty)\), so the correct answer is (C).

Step-by-step solution

Understanding the definition of an even function

A function f is considered even if for every x in the domain of f, the following property holds: f(x) = f(-x). Check if g(u) satisfies this property.

Testing if g(u) is an even function

Substitute -u into g to see if g(-u) equals g(u), i.e., check if \( g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2} \) is equal to \( g(u) = 2 \tan^{-1}(e^{u}) - \frac{\pi}{2} \).

Understanding the definition of an odd function

A function f is considered odd if for every x in the domain of f, the following property holds: f(-x) = -f(x). Check if g(u) satisfies this property.

Testing if g(u) is an odd function

Verify if \( g(-u)= -g(u) \) by simplifying \( -g(u) = -\left(2 \tan^{-1}(e^{u}) - \frac{\pi}{2}\right) \) and comparing it with \( g(-u) \).

Analyzing the monotonicity of g(u)

To test if g(u) is strictly increasing or decreasing, find its derivative \( g'(u) \) and determine the sign of \( g'(u) \) over the interval \( (-\infty, \infty) \).

Computing the derivative of g(u)

Differentiate \( g(u) \) with respect to u to get \( g'(u) = \frac{2e^u}{1+(e^u)^2} \).

Sign analysis of the derivative g'(u)

Since the exponent function \( e^u \) and the sum of exponent functions in the denominator are always positive, the derivative \( g'(u) \) is always positive for all u, which means g(u) is strictly increasing over \( (-\infty, \infty) \).

Key concepts

Even and Odd Functions

In understanding the nature of functions, categorizing them as even or odd can be incredibly useful. An even function is one for which the equation f(x) = f(-x) holds true for all x within the function's domain. This symmetry means that if you were to fold the graph of the function over the y-axis, it would align perfectly with itself, much like the letter 'M'. Popular examples of even functions include x^2 or cos(x).

On the other hand, an odd function is described by the equation f(-x) = -f(x), indicating that the graph of the function is symmetric with respect to the origin (the point (0,0)). Flipping it across both axes would bring you back the original graph. Typical instances of odd functions are x^3 and sin(x). When the function g(u) in our exercise is examined, it does not fulfil the criteria for either even or odd functions, meaning it's neither.

Differential Calculus

Differential calculus is a cornerstone of calculus that deals with change. At its heart lies the concept of the derivative, which measures how a function's output value changes in response to changes in its input value. Speaking in more tangible terms, it tells us the slope of the tangent line to the curve of a function at any given point.

The process involves several rules and techniques to maneuver through various types of functions, such as the power rule, product rule, and chain rule, to name a few. In the context of our problem, the derivative of g(u), denoted as g'(u), helps us determine the function's monotonicity - in simpler words, whether the function is increasing or decreasing, and how swiftly this change occurs.

Increasing and Decreasing Functions

A function can be described as increasing or decreasing based on whether its output values go up or down as the input values get larger. More precisely, a function is increasing if, for any two inputs x1 and x2 where x1 < x2, the output satisfies f(x1) ≤ f(x2). If the inequality is strict (f(x1) < f(x2)), we say the function is strictly increasing. This suggests that as we move along the graph from left to right, it rises upward.

Conversely, the function is labeled as decreasing when the output values drop as the input values climb. In the problem at hand, after computing the derivative g'(u), we find that it is always positive, which reveals that the function g(u) is strictly increasing throughout its entire domain. There's no heading downhill for this function – it's all uphill from any point!