Problem 40 - Essential Calculus Early Transcendentals

Question

Suppose \(f^{-1}\) is the inverse function of a differentiable function \(f\) and let \(G(x)=1 / f^{-1}(x)\) . If \(f(3)=2\) and \(f^{\prime}(3)=\frac{1}{y}\) , find \(G(2)\) .

Step-by-Step Solution

Answer

\(G(2) = \frac{1}{3}\).

Step 1: Identify the Relationship

Since we need to find \(G(2)\), start by identifying the relevant inverse relationship. Given that \(f(3) = 2\), it follows that \(f^{-1}(2) = 3\). This is because the inverse function \(f^{-1}\) reverses the input and output of \(f\).

Step 2: Differentiate the Inverse Function

To utilize the inverse relationship, differentiate it with respect to \(x\). By the chain rule and the fact that \(f(f^{-1}(x)) = x\), we have \(f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1\).

Step 3: Solve for the Derivative of the Inverse

Isolate \((f^{-1})'(x)\) from the equation derived in Step 2: \[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.\]

Step 4: Evaluate at the Specific Point

Substitute the specific values into the equation. Since \(f^{-1}(2) = 3\) and \(f'(3) = \frac{1}{y}\), we find:\[(f^{-1})'(2) = \frac{1}{f'(3)} = y.\]

Step 5: Find the Value of G(x)

Now compute \(G(2)\) using its definition: \[G(x) = \frac{1}{f^{-1}(x)}.\] For \(x = 2\), since \(f^{-1}(2) = 3\), it follows that: \[G(2) = \frac{1}{3}.\]

Key Concepts

Understanding Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. The derivative measures how a function's output changes as its input changes incrementally. Intuitively, it represents the slope of the function's curve at any given point, allowing us to understand how the function behaves. Differentiation is crucial when analyzing the rate of change, such as velocity in physics or growth rates in economics. It helps in identifying intervals where the function is increasing or decreasing.

Exploring the Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. When you have a function within another function, like \(f(g(x))\), the chain rule allows us to find its derivative efficiently.The chain rule states that if a function \(y\) depends on \(u\), and \(u\) depends on \(x\), then the derivative of \(y\) with respect to \(x\) is the product of the derivative of \(y\) with respect to \(u\) and \(u\) with respect to \(x\):

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

In the context of finding derivatives of inverses, the chain rule is often employed to manage the nested nature of functions effectively.

Delving into Derivatives

Derivatives are often denoted by \(f'(x)\) or \(\frac{dy}{dx}\), and they indicate how fast a function is changing at a particular point. The derivative is obtained by applying differentiation to the function. In essence, the derivative is a function itself that gives the rate of change at every point in the domain of the original function.Derivatives come in handy when looking to find areas like the slope of a curve at a point, describing the linear approximation of the function at that point, or solving optimization problems.

Inverse Relationships Explained

Inverse relationships in mathematics occur when one function undoes the effect of another function. If \(f(x)\) and \(f^{-1}(x)\) are functions, then \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). This property allows us to swap inputs and outputs to solve equations involving functions and their inverses.Understanding inverse relationships is vital when determining values like \(f^{-1}(x)\), as it reveals how changes in \(x\) affect the original function. Inverse functions form the backbone of operations like finding \(G(x) = \frac{1}{f^{-1}(x)}\), and understanding them is crucial for mastering calculus.