Problem 42 - A Textbook of Differential Calculus

Question

Which of the following pairs of function are identical? (a) \(f(x)=e^{\ln \sec ^{-1} x}\) and \(g(x)=\sec ^{-1} x\) (b) \(f(x)=\tan \left(\tan ^{-1} x\right)\) and \(g(x)=\cot \left(\cot ^{-1} x\right)\) (c) \(f(x)=\operatorname{sgn}(x)\) and \(g(x)=\operatorname{sgn}(\operatorname{sgn}(x))\) (d) \(f(x)=\cot ^{2} x \cdot \cos ^{2} x\) and \(g(x)=\cot ^{2} x-\cos ^{2} x\)

Step-by-Step Solution

Answer

Pairs (a), (b), and (c) are identical; pair (d) is not.

Step 1: Simplify Functions in Pair (a)

Evaluate the given functions in the first pair. For the first function: \[ f(x) = e^{\ln \sec^{-1} x} \]Use the property of exponentials and logarithms: \( e^{\ln a} = a \). Therefore, \[ f(x) = \sec^{-1} x \].Since \( g(x) = \sec^{-1} x \), we have \( f(x) = g(x) \). The functions are identical.

Step 2: Simplify Functions in Pair (b)

Evaluate the functions in the second pair.For \( f(x) = \tan (\tan^{-1} x) \), by the property of inverse functions, we get \( f(x) = x \).For \( g(x) = \cot (\cot^{-1} x) \), similarly, we find that \( g(x) = x \). Thus, \( f(x) = g(x) \). The functions are identical.

Step 3: Simplify Functions in Pair (c)

Consider the third pair of functions.For \( f(x) = \operatorname{sgn}(x) \), the function returns the sign of \( x \).For \( g(x) = \operatorname{sgn}(\operatorname{sgn}(x)) \), since \( \operatorname{sgn}(x) \) is 1, 0, or -1, the second application also results in the same value. Hence, \( f(x) = g(x) \). The functions are identical.

Step 4: Simplify Functions in Pair (d)

Evaluate the functions in the fourth pair.For \( f(x) = \cot^2 x \cdot \cos^2 x \), we can rewrite using identities: \( \cot x = \frac{1}{\tan x} \) and \( \cos^2 x + \sin^2 x = 1 \), simplifying gives a different result than \( \cot^2 x - \cos^2 x \). Therefore, \( f(x) eq g(x) \); the functions are not identical.

Step 5: Conclusion: Identify Identical Function Pairs

Pairs (a), (b), and (c) have functions that simplify to be identical, while pair (d) does not.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics with expressions that involve exponents. In the exercise, we encounter expressions like \( e^{\ln a} \), which exemplify a critical property of exponentials and logarithms.

To understand this property, remember that \( e \) is the base of the natural logarithm, approximately equal to 2.718. The exponential and logarithm functions are inverses, meaning \( e^{\ln a} = a \). This simplification is helpful in many mathematical problems.

In Pair (a) from the problem, this property allowed us to simplify \( f(x) = e^{\ln \sec^{-1} x} \) to \( f(x) = \sec^{-1} x \), proving the two functions were identical. Understanding these properties can greatly assist in solving similar problems involving exponential expressions.

Signum Function

The signum function, usually denoted as \( \operatorname{sgn}(x) \), plays a unique role in mathematics by extracting the sign of a number.

It returns 1 for positive numbers, -1 for negative numbers, and 0 if the number is zero. In essence, it simplifies a value to its basic sign characteristic, making it a fascinating tool in various fields.

In Pair (c) from the exercise, both functions \( f(x) \) and \( g(x) \) revolve around the signum function. When applied twice, \( \operatorname{sgn}(\operatorname{sgn}(x)) \), the result remains unchanged because it only depends on the sign of \( x \) itself, whether it is -1, 0, or 1. Therefore, \( f(x) \) and \( g(x) \) are identical as they yield the same output for any input \( x \). This property demonstrates the simplicity and utility of the signum function.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. These identities simplify complex trigonometric expressions and are indispensable in various mathematical computations.

Common identities include the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), and the cotangent identity, \( \cot x = \frac{1}{\tan x} \).

In the exercise's Pair (d), these identities were used to compare \( f(x) = \cot^2 x \cdot \cos^2 x \) and \( g(x) = \cot^2 x - \cos^2 x \). By employing the identities, we noticed that the two expressions were not equivalent since their simplifications resulted in different values.

Trigonometric identities like these are critical for transforming functions into more manageable forms, thus making it easier to evaluate their equivalency or solve equations.