Problem 5 - Calculus Early Transcendentals

Question

Suppose \(f\) is a one-to-one function with \(f(2)=8\) and \(f^{\prime}(2)=4 .\) What is the value of \(\left(f^{-1}\right)^{\prime}(8) ?\)

Step-by-Step Solution

Answer

Answer: \((f^{-1})^{\prime}(8) = \frac{1}{4}\)

Step 1: Write down the given information

We know that \(f(2)=8\), \(f^{\prime}(2)=4\), and we seek the value of \((f^{-1})^{\prime}(8)\).

Step 2: Apply the formula for the derivative of an inverse function

To find the derivative of the inverse function \(f^{-1}(x)\), we use the formula \((f^{-1})'(x) = \frac{1}{f' (f^{-1}(x))}\).

Step 3: Evaluate the inverse function inside the derivative

We want to find the value of \((f^{-1})^{\prime}(8)\). Therefore, we need to evaluate \(f^{-1}(8)\). Since \(f(2) = 8\), we know that \(f^{-1}(8) = 2\).

Step 4: Substitute and find the derivative of the inverse function

Now, using the formula from Step 2, we plug in the values we found: \((f^{-1})^{\prime}(8) = \frac{1}{f' (f^{-1}(8))} = \frac{1}{f'(2)}\). Since we know that \(f^{\prime}(2) = 4\), we can plug in this value and find the derivative of the inverse function: \((f^{-1})^{\prime}(8) = \frac{1}{f'(2)} = \frac{1}{4}\). So, the value of \((f^{-1})^{\prime}(8) = \frac{1}{4}\).

Key Concepts

Derivatives

Derivatives are a core concept in calculus, often nicknamed the cornerstone of calculus. Derivatives represent the rate of change or the slope of a function at any given point. They tell us how a function behaves locally, like how fast a car is going at exactly 3 seconds.

In the case of the exercise above, the derivative of function \(f\) at \(x = 2\) is 4, denoted as \(f^{\prime}(2) = 4\). This means that at \(x = 2\), the function \(f\) is changing at a rate of 4 units per unit of \(x\). In simpler terms, it's like saying that a car is moving at 4 meters per second at that moment.

Knowing how to find and interpret derivatives helps in understanding the behavior of functions, optimizing various scenarios, and solving many mathematical problems related to change.

Inverse Functions

Inverse functions essentially reverse the effects of the original function. If a function \(f\) takes a value \(x\) to \(y\), then its inverse function \(f^{-1}\) will take \(y\) back to \(x\).

For an inverse function to exist, the original function needs to be one-to-one (bijective), meaning it pairs each \(x\) with exactly one \(y\) and vice versa. The example function \(f\) is such that \(f(2) = 8\) and therefore, \(f^{-1}(8) = 2\).

When differentiating inverse functions, a special formula is used:

• \((f^{-1})'(x) = \frac{1}{f' (f^{-1}(x))}\)

This formula tells us that the derivative of the inverse function at a point is the reciprocal of the derivative of the original function, evaluated at the inverse function's input. Understanding this relationship is crucial in both pure and applied mathematics.

Calculus

Calculus encompasses two major branches: differentiation and integration. In this context, we focus on the differentiation aspect, which involves derivatives and their applications. Calculus allows us to study how functions change, handle concepts of motion, and solve equations that model real-world problems.

Solving exercises about inverse functions and their derivatives, like the one presented, requires a good grasp of calculus concepts. The exercise involves understanding how to manipulate both functions and their inverses — a process made simpler with calculus knowledge.

Calculus is essential because it provides the tools needed for precise calculation of limits, derivatives, and integrals, addressing how quantities accumulate and change. This fundamental branch of mathematics supports sciences, engineering, economics, and beyond.