Problem 17 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}\), then \(f^{-1}(x)\) is (a) \(\frac{1}{2} \log _{2} \frac{x}{1-x}\) (b) \(\frac{1}{2} \log _{2} \frac{1+x}{1-x}\) c) \(\frac{1}{2} \log _{2} \frac{1+x}{x}\) (d) \(\frac{1}{2} \log _{2} \frac{2+x}{2-x}\)

Step-by-Step Solution

Answer

The inverse function is (b) \( \frac{1}{2} \log_{2} \frac{1+x}{1-x} \).

Step 1: Analyze the Given Function

The function given is \( f(x) = \frac{2^{x}-2^{-x}}{2^{x}+2^{-x}} \). Our goal is to find the inverse function \( f^{-1}(x) \). This means we want to express \( x \) in terms of \( y \), where \( y = f(x) \).

Step 2: Substitute and Simplify

Let \( y = f(x) = \frac{2^{x} - 2^{-x}}{2^{x} + 2^{-x}} \). We need to express \( x \) in terms of \( y \). Rearranging the equation gives: \( y(2^{x} + 2^{-x}) = 2^{x} - 2^{-x} \).

Step 3: Rearrange the Terms

We have \( y \cdot 2^{x} + y \cdot 2^{-x} = 2^{x} - 2^{-x} \). Rearrange this to get: \( 2^{x}(1-y) = 2^{-x}(1+y) \).

Step 4: Solve for Exponents

By cross-multiplying, we get \( 2^{2x} = \frac{1+y}{1-y} \). Solving for \( 2^{x} \), we take the square root of both sides: \( 2^{x} = \sqrt{\frac{1+y}{1-y}} \). This simplifies to \( 2^{x} = \left(\frac{1+y}{1-y}\right)^{1/2} \).

Step 5: Use Logarithms to Solve for x

Take the logarithm base 2 of both sides: \( x = \frac{1}{2} \log_{2}\left(\frac{1+y}{1-y}\right) \). This forms the inverse function.

Key Concepts

Understanding Logarithms

Logarithms serve as the inverse operations to exponential functions. In the previous solution, we used logarithms to solve for the variable after expressing the exponential form of the function. If you have an equation of the form \( 2^x = a \), taking the logarithm base 2 of both sides gives \( x = \log_{2}(a) \). This operation helps us to isolate the exponent, making it easier to solve for it.

• Logarithms convert multiplication into addition, which simplifies many algebraic manipulations.
• They provide a way to transform complex equations into forms that are easier to manage.
• When dealing with inverse functions, using logarithms is key to "undoing" what the exponential function has done.

In our solution, we used the property of logarithms to determine the inverse expression, assisting us to find \( f^{-1}(x) \) by converting an exponential statement back to a logarithmic form.

Expanding on Exponential Functions

Exponential functions, like \( 2^x \), are powerful mathematical expressions. They involve variables being raised as exponents. The familiar base 2 is prevalent here, representing processes that change at constant proportional rates. The original function \( f(x) = \frac{2^{x} - 2^{-x}}{2^{x} + 2^{-x}} \) used these exponential terms extensively. Let's dissect the core concept:

• Exponentials grow or decay rapidly, making them useful to describe phenomena like population growth or nuclear decay.
• When working with inverse functions, understanding exponential growth helps to anticipate how logarithms will unwind the process.
• This rapid change is illustrated more deeply by the presence of \( 2^{-x} \), seen in the given function, which decreases as \( x \) increases. This forms the basic structure of symmetric hyperbolic forms with exponential components.

Recognizing the role of exponentials in the function was crucial in breaking down and forming the correct inverse by later using its logarithmic counterpart.

Function Transformation

Function transformation involves changing a function’s appearance without altering its fundamental properties. Inverse functions are a form of transformation, where you essentially 'flip' inputs with outputs. In the exercise solution, transforming the function from its original to an inverse involved deliberate steps:

• Swapping \( x \) with \( y \), therefore seeking an expression for \( x \) in terms of \( y \).
• The rearranging of terms helps visualize transformation necessary to isolate and solve for \( x \).
• This comprehension of function transformation is key when flipping relationships, ensuring that each step logically leads to the eventual goal of the inverse function.

By understanding these changes, we see transformations aren't just about visual manipulations but also about preserving relationships in the new expression. The ultimate goal is to logically approach the inverse through simplified algebraic rearrangements.