Problem 42 - Thomas Calculus

Question

$$\begin{array}{l}{\text { Let } f(x)=x^{2}-4 x-5, x>2 . \text { Find the value of } d f^{-1} / d x \text { at the }} \\ {\text { point } x=0=f(5) .}\end{array}$$

Step-by-Step Solution

Answer

\(\frac{1}{6}\)

Step 1: Clarify the Task

We need to find the derivative of the inverse function of \(f(x)\) at a given point \(x = 0\), where \(x=0\) is the output of \(f(x)\) when \(x=f^{-1}(0)\).

Step 2: Confirm the Given Point

Since \(f(x) = x^2 - 4x - 5\), and we are given \(f(5) = 0\), we conclude that \(f^{-1}(0) = 5\). Thus, we need to find \(\left. \frac{d f^{-1}}{dx} \right|_{x=0}\).

Step 3: Use the Derivative of Inverse Function Formula

The formula for the derivative of a function's inverse is given by \(\frac{d f^{-1}}{dx} = \frac{1}{f'(f^{-1}(x))}\). We need to calculate \(f'(x)\) first.

Step 4: Differentiate \(f(x)\)

Differentiate \(f(x) = x^2 - 4x - 5\) with respect to \(x\). This gives:\[ f'(x) = 2x - 4 \].

Step 5: Substitute the Inverse Point into the Derivative

Since we found that \(f^{-1}(0) = 5\), substitute \(x = 5\) into \(f'(x)\) to get:\[ f'(5) = 2(5) - 4 = 10 - 4 = 6 \].

Step 6: Calculate the Derivative of the Inverse

Using the inverse derivative formula, substitute into \(\frac{1}{f'(f^{-1}(x))}\):\[ \frac{d f^{-1}}{dx} = \frac{1}{6} \].

Step 7: Verify and Conclude

Verify all calculations are correct and consistent with principles. Thus, \( \left. \frac{d f^{-1}}{dx} \right|_{x=0} = \frac{1}{6} \).

Key Concepts

Inverse Functions

Understanding inverse functions is vital in calculus problems, particularly when dealing with differentiation. An inverse function essentially "undoes" the action of the original function. If you think of a function as a machine that takes an input, processes it, and delivers an output, the inverse function takes that output and returns it to the original input.

For example, if you have a function \(f\) such that \(f(x) = y\), the inverse function, denoted \(f^{-1}(y)\), would then give you back \(x\) when you input \(y\).

• the function \( f(x) = x^2 - 4x - 5 \) takes a number \( x > 2 \) and computes a value;
• the inverse \( f^{-1}(x) \) will reverse this result to the original input.

To find these inverse functions, solving the equation \(f(x)=y\) for \(x\) is required. Recognizing when a function has an inverse, logically follows from understanding its one-to-one nature, particularly crucial for tasks like differentiation of these functions.

Function Differentiation

Function differentiation is the process of finding the derivative, which measures how a function changes as its input changes. In the context of inverse functions, it's crucial to understand how differentiating the original function effects its inverse.

Let's examine the original function \( f(x) = x^2 - 4x - 5 \). To differentiate it, we need to find its derivative with respect to \( x \). This derivative, denoted \( f'(x) \), is computed using standard differentiation rules, leading to \( f'(x) = 2x - 4 \).

• The expression \( 2x - 4 \) indicates the slope of the tangent line at any point \( x \).
• Positive or negative, this derivative reveals increasing or decreasing tendencies of the function at various intervals.

This slope is pivotal when determining the behavior of not just the function, but also its inverse, as it guides the inverse derivative calculation.

Calculus Problems

Calculus involves solving various types of problems using differentiation and integration. One of the more challenging tasks can be finding the derivative of an inverse function. Here, the derivative of the inverse is not something that's directly computed like in typical scenarios. Instead, it involves a formula:

\[ \frac{d f^{-1}}{dx} = \frac{1}{f'(f^{-1}(x))} \]

Using this, we substitute \( f'(f^{-1}(0)) \). Since we have calculated \( f^{-1}(0) = 5 \), inserting \( x = 5 \) into \( f'(x) \) yields \( f'(5)=6 \).

• This formula essentially flips the derivative, indicating a reciprocal relationship between \( f' \) and \( f^{-1}' \).
• It underscores the unique interplay between a function and its inverse in calculus.

Solutions like these highlight the depth and intricacy of calculus problems, pushing learners to apply concepts holistically.

Mathematical Concepts

Grasping the mathematical concepts underlying these calculations is crucial for solving inverse function derivatives effectively. This involves combining various ideas such as function behavior, slope, and reciprocal relationships.

At the heart lies understanding how functions transform and the nuanced interrelations with their inverses.

• Recognizing patterns and generalizing seems abstract, but greatly simplifies complex problem-solving.
• Concepts like one-to-one functions, necessary for existing inverses, prominence in these exercises highlight their practical utility.

Familiarity with such concepts not only aids in mastering inverse function derivatives but also enriches one's broader mathematical knowledge, crucial for advanced topics in calculus and beyond.