Problem 123 - Precalculus

Question

We also prove in calculus that the derivative of the inverse function \(f^{-1}\) is given by \(\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}\). Given \(f(x)=e^{x},\) find a. \(f^{-1}(x)\) b. \(\left(f^{-1}\right)^{\prime}(x)\)

Step-by-Step Solution

Answer

a. \(f^{-1}(x) = \ln(x)\); b. \(\left(f^{-1}\right)'(x) = \frac{1}{x}\).

Step 1: Identify the Function

The function given is \(f(x) = e^x\). The task is to find its inverse \(f^{-1}(x)\).

Step 2: Find the Inverse Function

To find the inverse, we set \(y = e^x\). Swap \(x\) and \(y\) to get \(x = e^y\). Solving for \(y\), we take the natural logarithm on both sides: \(y = \ln(x)\). Thus, the inverse function is \(f^{-1}(x) = \ln(x)\).

Step 3: Differentiate the Inverse Function

To find \(\left(f^{-1}\right)'(x)\), we use the derivative of \(\ln(x)\), which is \(\frac{1}{x}\). Therefore, \(\left(f^{-1}\right)'(x) = \frac{1}{x}\).

Step 4: Verify Formula for Derivative of Inverse

The formula for the derivative of an inverse function is \( \left(f^{-1}\right)^{\prime}(x) = \frac{1}{f^{\prime}(f^{-1}(x))} \). We have \(f(x) = e^x\) so \(f'(x) = e^x\). Substitute \(f^{-1}(x) = \ln(x)\) into the derivative: \(\frac{1}{f^{\prime}(\ln(x))} = \frac{1}{e^{\ln(x)}} = \frac{1}{x} \), which matches our previous result.

Key Concepts

Derivative of Inverse

The concept of the derivative of an inverse function is a fascinating topic in calculus. When you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), knowing how they relate through derivatives is powerful. The derivative of an inverse function, denoted as \( (f^{-1})'(x) \), can be found using the formula:

• \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \).

This formula provides a neat way to determine the rate of change of the inverse function by utilizing the derivative of the original function.
To understand this, think of it as recovering the slope of the original function through its inverse at a particular point.The derivative formula implies symbiosis between \( f(x) \) and \( f^{-1}(x) \). If you know \( f(x) = e^x \), its inverse needs to be determined first before applying this derivative formula. This shows how derivatives can not only tell us about rates of change but retain their properties even when the functions are flipped upside-down!

Natural Logarithm

The natural logarithm is an essential concept in mathematics, especially when working with exponential functions. Referred to as \( \ln(x) \), it functions as the inverse of the exponential function \( e^x \). Here’s a quick breakdown:

• \( \ln(e^x) = x \)
• Conversely, \( e^{\ln(x)} = x \)

This relationship underlines the seamless transition between exponential growth and logarithmic analysis.
In the scenario where you are tasked with finding the inverse of an exponential function, taking the natural logarithm is your method of unraveling that exponential wrapping.A good example of this happens when finding the inverse of the function \( f(x) = e^x \). You swap variables to solve for \( y \), and introduce \( \ln(x) \) because its capability to revert the effects of exponentiation on \( x \).This introduces the natural logarithm as a vital mathematical tool for simplifying complex problems that involve exponential calculations.

Exponential Function

Exponential functions, especially \( e^x \), are crucial in various mathematical contexts. They represent continuous growth processes and naturally occur in diverse fields like finance, physics, and biology.
The base \( e \) is a mathematical constant approximately equal to 2.718. Its unique properties make functions like \( f(x) = e^x \) particularly interesting because:

• The derivative of \( e^x \) is \( e^x \) itself, enabling smooth and elegant calculus operations.
• Inverse problems become structured, as this property informs how we calculate derivatives of its inverse function.

The beauty of the exponential function lies in its inherent growth characteristics; it models compound interest, population growth, and radioactive decay.
Its version with \( e \) is most useful in natural processes due to its constant derivation retaining the same base, leading to straightforward solutions in translating continuous change into mathematical language.