Problem 62 - A Textbook of Differential Calculus

Question

\(f: R \rightarrow R \quad\) and \(\quad f(x)=a_{1} x+a_{3} x^{3}+a_{5} x^{5}+\ldots+a_{2 n+1} x^{2 n+1}-\cot ^{-1} x\) where \(0

Step-by-Step Solution

Answer

The function \( f(x) \) is one-one onto (option c).

Step 1: Analyze Function Type

Examine the given function \( f(x) = a_{1}x + a_{3}x^{3} + a_{5}x^{5} + \ldots + a_{2n+1}x^{2n+1} - \cot^{-1}x \). It consists of an odd degree polynomial and the function \( \cot^{-1}x \), both of which are continuous and differentiable.

Step 2: Evaluate Monotonicity

To determine if the function is one-one (injective), check the first derivative: \[ f'(x) = a_1 + 3a_3x^2 + 5a_5x^4 + \ldots + (2n+1)a_{2n+1}x^{2n} + \frac{1}{1+x^2} \]Given that all \( a_{i} > 0 \) are increasing, and \( \frac{1}{1+x^2} > 0 \), \( f'(x) > 0 \) for all \( x \). Hence, \( f(x) \) is strictly increasing, indicating the function is one-one.

Step 3: Determine Onto Property

To verify if the function is onto (surjective), consider the behavior as \( x \to \pm \infty \). Since the highest degree term \( a_{2n+1} x^{2n+1} \) dominates as \( x \to \pm \infty \), the polynomial ranges from \( -\infty \to \infty \) and \( \cot^{-1}x \to 0 \) in these limits. Thus, \( f(x) \) covers all real values \((-\infty, \infty)\), making it onto.

Step 4: Conclusion on Type

Since \( f(x) \) is both one-one and onto, it is a bijection.

Key Concepts

Understanding Injective Functions

Injective functions, often called one-to-one functions, are a key concept in mathematics. They are functions where every element of the function's domain (input) maps to a unique element in the codomain (output). This means no two different inputs in the domain will map to the same output in the codomain. To determine if a function is injective, we often check if the function is strictly increasing or strictly decreasing by looking at its derivative.

For example, when examining a function like:

• \( f(x) = a_{1}x + a_{3}x^{3} + a_{5}x^{5} + \ldots + a_{2n+1}x^{2n+1} - \cot^{-1}x \)

We can check its derivative, \( f'(x) \). In our original problem, \( f'(x) = a_1 + 3a_3x^2 + 5a_5x^4 + \ldots + (2n+1)a_{2n+1}x^{2n} + \frac{1}{1+x^2} \). Since all coefficients \( a_i \) are positive and \( \frac{1}{1+x^2} \) is positive, \( f'(x) > 0 \). Therefore, this function is strictly increasing and injective by nature.

This strict monotonicity leads to the conclusion that each input has a distinct output, confirming the injective nature of the function.

Exploring Surjective Functions

Surjective functions are another essential type of function, known as onto functions. A function is surjective if every element in the codomain is an image of at least one element from the domain. In simpler terms, the entire codomain is covered by the function.

To determine if a function is surjective, consider the behavior of the function as its input goes to positive or negative infinity. For the function we examined:

• The highest degree term \( a_{2n+1}x^{2n+1} \) dominates as \( x \to \infty \) or \( x \to -\infty \).

This term makes the polynomial stretch across all possible real numbers, from \(-\infty\) to \(+\infty\). Moreover, as \( x \to \infty \) or \( x \to -\infty \), \( \cot^{-1}x \to 0 \), allowing the function to span all real values.

Therefore, the given function is surjective as it maps the real numbers (domain) onto the entire set of real numbers (codomain), achieving full coverage.

Introduction to Differentiable Functions

Differentiable functions are a class of functions that are smooth and have a well-defined derivative at every point in their domain. The concept of differentiability is fundamental in calculus and provides insight into how functions behave, such as whether they are increasing or decreasing.

A differentiable function must first be continuous; however, not all continuous functions are differentiable. The differentiability of a function like:

• \( f(x) = a_{1}x + a_{3}x^{3} + a_{5}x^{5} + \ldots + a_{2n+1}x^{2n+1} - \cot^{-1}x \)

can be understood by examining its derivative, \( f'(x) \), as we did in our analysis. Here, the derivative exists across all points because both polynomial components and the arc-cotangent are smooth.

This smooth nature means we can explore the function's behavior using calculus tools, such as finding its slope at any given point. Hence, our function is not only continuous but also differentiable, allowing us to utilize its first and higher-order derivatives for deeper insights into its character.