Problem 39 - Essential Calculus Early Transcendentals

Question

Suppose \(f^{-1}\) is the inverse function of a differentiable function \(f\) and \(f(4)=5, f^{\prime}(4)=\frac{2}{3} .\) Find \(\left(f^{-1}\right)^{\prime}(5)\)

Step-by-Step Solution

Answer

The derivative \((f^{-1})'(5) = \frac{3}{2}\).

Step 1: Understanding the Problem

We are given a function \(f(x)\) with \(f(4) = 5\) and its derivative \(f'(4) = \frac{2}{3}\). We need to find the derivative \((f^{-1})'(5)\) of the inverse function at \(x = 5\).

Step 2: Identify the Relationship between Function and Its Inverse

We know that if \(y = f(x)\), then the inverse function \(f^{-1}(y) = x\). Since \(f(4) = 5\), it means \(f^{-1}(5) = 4\).

Step 3: Apply the Inverse Function Derivative Formula

The derivative of an inverse function can be found using the formula: \((f^{-1})'(y) = \frac{1}{f'(x)}\), where \(y = f(x)\). In our case, \(y = 5\) and when \(y=5\), \(x=4\).

Step 4: Substitute the Known Values into the Derivative Formula

Using the formula \((f^{-1})'(5) = \frac{1}{f'(4)}\), substitute \(f'(4) = \frac{2}{3}\) into it: \((f^{-1})'(5) = \frac{1}{\frac{2}{3}}\).

Step 5: Simplify the Expression

Simplify the expression \(\frac{1}{\frac{2}{3}}\) by flipping the fraction: \(\frac{3}{2}\). Thus, \((f^{-1})'(5) = \frac{3}{2}\).

Key Concepts

Differentiable Function

A differentiable function is a type of mathematical function that possesses a derivative at every point in its domain. This means the function has a well-defined tangent line at each point, allowing us to measure how the function changes at that point. The derivative of a function, denoted as \( f'(x) \), essentially provides the rate at which the function's value is changing with respect to \( x \).

• To check if a function is differentiable, it must be smooth without any sharp turns or corners.
• The existence of a derivative implies that the function is continuous. However, continuity alone does not guarantee differentiability.

Differentiability is a crucial concept as it underpins many aspects of calculus. It allows for the application of tools such as the derivative to analyze and solve real-world problems related to rates and trends. In this specific problem, the function \( f \) is differentiable, and we know its derivative value at a point: \( f'(4) = \frac{2}{3} \). This derivative will be instrumental when working with its inverse function.

Inverse Function Derivative Formula

The inverse function derivative formula is a powerful tool in calculus. It gives us the derivative of an inverse function without directly computing the inverse function itself. The formula is given by:\[(f^{-1})'(y) = \frac{1}{f'(x)}\]Here, \( y = f(x) \) and \( x = f^{-1}(y) \). This formula is derived from the relationship between a function and its inverse. It's particularly useful in problems where finding the inverse function explicitly can be cumbersome or impossible.In our exercise, we applied this formula by first understanding that \( y = 5 \) (from \( f(4) = 5 \)), so the inverse at \( y = 5 \) gives \( x = 4 \). Then, using the known derivative \( f'(4) = \frac{2}{3} \), the inverse derivative becomes:\[(f^{-1})'(5) = \frac{1}{f'(4)} = \frac{1}{\frac{2}{3}} = \frac{3}{2}\]This simplified expression shows the rate at which the inverse function changes when \( y \) changes near 5.

Function and Inverse Relationship

Understanding the relationship between a function and its inverse is essential for grasping many calculus concepts. If you have a function \( f(x) \) with inverse \( f^{-1}(y) \), it means:

• For every pair \((x, y)\) where \( y = f(x) \), the inverse function satisfies \( x = f^{-1}(y) \).
• The function \( f \) maps inputs to outputs, while the inverse maps those outputs back to the original inputs.

In this problem, we see that if \( f(4) = 5 \), then the inverse relationship gives us \( f^{-1}(5) = 4 \). This relationship is visualized as swapping the roles of inputs and outputs. Therefore, inverting a function allows us to reverse the process of the function, retrieving the original input value from the output.This understanding is especially important for problems involving changes and rates, such as differentiable functions and their inverses. Derivatives of inverses help us analyze the dynamics of the original function’s output relative to changes in its input, connecting the geometric perspective of functions with algebraic manipulation.