Problem 124 - 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 \(f(x)=e^{2 x},\) find a. \(f^{-1}(x)\) b. \(\left(f^{-1}\right)^{\prime}(x)\)

Step-by-Step Solution

Answer

a. \( f^{-1}(x) = \frac{\ln(x)}{2} \); b. \( \left( f^{-1} \right)'(x) = \frac{1}{2x} \).

Step 1: Find the Inverse Function

The given function is \( f(x) = e^{2x} \). To find the inverse, we set \( y = e^{2x} \), which gives us \( x = \ln(y) / 2 \). Thus, the inverse function is \( f^{-1}(x) = \frac{\ln(x)}{2} \).

Step 2: Find the Derivative of the Original Function

The derivative of the original function \( f(x) = e^{2x} \) with respect to \( x \) is \( f'(x) = 2e^{2x} \). This will help us find the derivative of the inverse function.

Step 3: Apply the Derivative of the Inverse Function

We use the formula for the derivative of the inverse function: \( \left( f^{-1} \right)'(x) = \frac{1}{f'(f^{-1}(x))} \). Since \( f^{-1}(x) = \frac{\ln(x)}{2} \), substitute it into the derivative of \( f \): \[ f'(f^{-1}(x)) = f'\left(\frac{\ln(x)}{2}\right) = 2e^{2 \cdot \frac{\ln(x)}{2}} = 2x \].

Step 4: Simplify the Derivative of the Inverse Function

Substitute back into the formula: \[ \left( f^{-1} \right)'(x) = \frac{1}{2x} \]. This is the derivative of the inverse function.

Key Concepts

Derivatives

Derivatives measure the rate at which a function's output changes in response to a change in its input. They are a fundamental part of calculus and have applications in various fields such as physics, engineering, and economics. When dealing with inverse functions, the formula for the derivative of the inverse is particularly important: \[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \]This formula allows us to find the derivative of the inverse function using the derivative of the original function. It is important to understand how this formula is derived and applied. - To apply this formula, first find the inverse function.- Next, find the derivative of the original function.- Finally, substitute these into the formula to find the derivative of the inverse. In the given problem, for the function \( f(x) = e^{2x} \), we calculate its derivative \( f'(x) \), which is \( 2e^{2x} \), and utilize this derivative in calculating the derivative of its inverse. Understanding derivatives and their rules makes tackling such problems more approachable.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent. They are written in the form \( f(x) = a^x \) or, with the natural base, \( f(x) = e^x \). These functions model many natural phenomena, such as population growth and radioactive decay. - The base \( e \) is approximately equal to 2.71828 and is used widely in calculus due to its unique properties.- In exponential functions, the rate of change of the function is proportional to the value of the function itself. For the exercise above, the given function is \( f(x) = e^{2x} \), which is an exponential function where the variable \( x \) is multiplied by 2 before being raised as an exponent of \( e \). Understanding exponential functions helps in many areas, from solving differential equations to financial modeling.

Logarithmic Functions

A logarithmic function is the inverse of an exponential function. If \( f(x) = e^x \), then \( f^{-1}(x) = \ln(x) \). Logarithms reverse the operation of exponentiation and allow computation of the time necessary to reach a certain level of growth. - The natural logarithm, denoted as \( \ln \), is particularly important because it is the inverse of the exponential function with base \( e \).- Logarithmic functions transform multiplicative relationships into additive relationships, simplifying complex equations. In the original problem, we find the inverse of the exponential function \( f(x) = e^{2x} \). The inverse is \( f^{-1}(x) = \frac{\ln(x)}{2} \). Here, the logarithmic function transforms the exponential output back to its corresponding input value, adjusted by the factor dividing the logarithm. Logarithmic transformations are used extensively in data analysis and scientific calculations to linearize exponential relationships.