Problem 292 - Calculus Volume 1

Question

For the following exercises, use the given values to find \(\left(f^{-1}\right)^{\prime}(a).\) $$f(\sqrt{3})=\frac{1}{2}, f^{\prime}(\sqrt{3})=\frac{2}{3}, a=\frac{1}{2}$$

Step-by-Step Solution

Answer

\((f^{-1})'\left(\frac{1}{2}\right) = \frac{3}{2}.\)

Step 1: Understanding the Problem

We are given a function \( f \) with specific values and its derivative at a particular point. We need to find the derivative of the inverse function \( \, (f^{-1})^{\prime}(a) \, \), where \( a = \frac{1}{2} \).

Step 2: Recall the Formula for Inverse Derivative

The derivative of the inverse function can be found using the formula:\[(f^{-1})^{\prime}(a) = \frac{1}{f^{\prime}(f^{-1}(a))}\] This formula helps us find the slope of the inverse function at a point.

Step 3: Identifying \(f^{-1}(a)\)

Since \( a = \frac{1}{2} \), this value corresponds to \( f(x) \). From the problem, \( f(\sqrt{3}) = \frac{1}{2} \). Therefore, \( f^{-1}(\frac{1}{2}) = \sqrt{3} \).

Step 4: Substitute Values into the Formula

Substitute \( f^{-1}(\frac{1}{2}) = \sqrt{3} \) and \( f^{\prime}(\sqrt{3}) = \frac{2}{3} \) into the formula for the inverse function's derivative:\[(f^{-1})^{\prime}\left(\frac{1}{2}\right) = \frac{1}{f^{\prime}(\sqrt{3})} = \frac{1}{\frac{2}{3}}\]

Step 5: Calculate the Result

Perform the division to find the derivative of the inverse function:\[(f^{-1})^{\prime}\left(\frac{1}{2}\right) = \frac{3}{2}\]

Key Concepts

Inverse Function

In mathematics, an inverse function is a function that reverses another function. Imagine a function is a machine that takes an input and produces an output. An inverse function does the opposite - it takes that output and returns the original input. Let’s say you have a function, denoted as \(f\), and you input a number \(x\). The function gives you an output \(y\), meaning \(f(x) = y\). The inverse function, denoted as \(f^{-1}\), does the reverse. It takes \(y\) and gives you \(x\), so \(f^{-1}(y) = x\). For a function to have an inverse, it must be bijective. This means it is both one-to-one (each output is produced by one input) and onto (all possible outputs are used). When these conditions are met, the inverse function essentially "undoes" the function's effect.

Derivative Formula

The derivative of an inverse function is found using a specific formula, which is critical in calculus for understanding how rates of change are related between functions and their inverses. The formula is:\[(f^{-1})^{\prime}(a) = \frac{1}{f^{\prime}(f^{-1}(a))}\]This might look complex at first, but let's break it down. When we want to find the derivative of an inverse function at a point \(a\), we first determine \(f^{-1}(a)\), which is the "input" that would give \(a\) as an "output" through \(f\).After identifying \(f^{-1}(a)\), substitute it into \(f'\), the derivative of the original function. Essentially, you're asking: "At what rate is the original function changing at the point that corresponds to \(a\) on the inverse function?" By taking the reciprocal of this rate, you determine the rate of change (or slope) of the inverse function at \(a\). This reciprocal relationship is key in understanding how inverses operate.

Function Inversion

Function inversion involves identifying the input-output relationship between a function and its inverse. In simpler terms, you look at what input a function maps to an output and reverse these roles for the inverse function. This process can seem daunting but becomes intuitive with a bit of practice:

• Start with a clearly defined function \(f\).
• Identify pairs: Paring each output back to its original input, effectively reversing the function.
• Ensure the function is bijective (invertible) to proceed with inversion.

The highlighted example from the step-by-step solution illustrates this. Given that \(f(\sqrt{3}) = \frac{1}{2}\), the inverse function \(f^{-1}\) brings \(\frac{1}{2}\) back to \(\sqrt{3}\).Understanding this process is crucial for delving into more complex functions and plays a vital role in calculus, allowing for the exploration of inverse relationships and how they interact with derivative concepts.