Problem 291 - Calculus Volume 1

Question

Use the given values to find \(\left(f^{-1}\right)^{\prime}(a)\). $$ f\left(\frac{1}{3}\right)=-8, f^{\prime}\left(\frac{1}{3}\right)=2, a=-8 $$

Step-by-Step Solution

Answer

\( \left(f^{-1}\right)'(-8) = \frac{1}{2} \)

Step 1: Find the Relation for Inverse Derivative

The formula to find the derivative of an inverse function given \( f(x) \) is \( \left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))} \). This means you can calculate the derivative of the inverse function at \( a \) once you know \( f(x) \) and its derivative.

Step 2: Identify Given Values and Relations

From the problem, we know that \( f\left(\frac{1}{3}\right) = -8 \). This implies that if \( f(x) = -8 \), then \( x = \frac{1}{3} \). Therefore, \( f^{-1}(-8) = \frac{1}{3} \). Also, given that \( f'(\frac{1}{3}) = 2 \), use this to find \( \left(f^{-1}\right)'(-8) \).

Step 3: Apply Inverse Function Derivative Formula

Plug the known values into the inverse derivative formula: \[ \left(f^{-1}\right)'(-8) = \frac{1}{f'(f^{-1}(-8))} = \frac{1}{f'(\frac{1}{3})} \].

Step 4: Calculate the Derivative of the Inverse at a

Substitute \( f'(\frac{1}{3}) = 2 \) into the formula from the previous step: \[ \left(f^{-1}\right)'(-8) = \frac{1}{2} \].

Key Concepts

Calculus

Calculus is a branch of mathematics focused on the concepts of change, typically using derivatives and integrals to explore different problems. Understanding calculus is crucial for solving problems related to motion, growth, decay, and optimization.
- The derivative represents how a function changes as its input changes. It's the mathematical equivalent of finding the speed of a car by measuring how far it goes over a period of time. - Calculus relies largely on limits and infinitesimally small changes to compute values that aren't immediately obvious.
By using calculus, you can find rates of change or the area under a curve with precision, helping you explore the behaviors of different functions more deeply.
Understanding these concepts lays the foundation for dealing with more complex problems such as the inverses of functions and their derivatives.

Derivative of Inverse Functions

The derivative of an inverse function tells us how the inverse function changes. It's directly related to the derivative of the original function, measured at specific points. Knowing this derivative helps in many practical applications, such as finding the rate at which one variable affects another.
To find the derivative of an inverse function, you use this important formula:

• \( \left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))} \)

This formula reveals that the derivative of the inverse function at a given point \( a \) is the reciprocal of the derivative of the original function, calculated at the point where the original function equals \( a \).
For instance, in the solution provided, we find \( f^{-1}(-8) \) by identifying that \( f(\frac{1}{3}) = -8 \) and then compute the inverse derivative using the given derivative at this point: \( \frac{1}{2} \). This demonstrates how these concepts are not only theoretical but applicable in finding precise values in real-world problems.

Function Inverses

Function inverses reverse the operation done by the function. Whereas a function maps inputs to outputs, its inverse maps outputs back to the original inputs. Understanding function inverses is crucial in various fields like engineering and physics, where reversing processes is a common task.
For two functions, \(f\) and \(f^{-1}\), to be inverses, the following must be true:

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

In the exercise solution, knowing that \(f(\frac{1}{3}) = -8\) helped us determine that \(f^{-1}(-8) = \frac{1}{3}\), demonstrating how inverting the function provides a way to retrace our steps back to the original input.
Working with function inverses requires careful manipulation of equations and an understanding of how these inverses relate back to their original functions. Studying inverses gives insights into functions' behavior and increases problem-solving flexibility.