Problem 8 - Calculus Single Variable

Question

Assume that \(f: \mathbb{R} \rightarrow \mathbb{R}\) is invertible and differentiable. Compute \(\left(f^{-1}\right)^{\prime}(4)\) from the given information. $$ f(0)=4, f^{\prime}(s)=2 e^{2 s}+e^{-s} $$

Step-by-Step Solution

Answer

\((f^{-1})'(4) = \frac{1}{3}\).

Step 1: Understand Inverse Derivative Formula

To find the derivative of an inverse function at a given point, we can use the formula: \[ (f^{-1})'(b) = \frac{1}{f'(a)} \] where \(f(a) = b\). In this problem, we are given \(f(0) = 4\), which means \(a = 0\) and \(b = 4\). Our task is to find \((f^{-1})'(4)\).

Step 2: Calculate the Derivative at the Given Point

Use the given derivative \(f'(s) = 2e^{2s} + e^{-s}\) to find \(f'(0)\). Substitute \(s = 0\) into the expression: \[ f'(0) = 2e^{0} + e^{0} \] Simplifying gives \(f'(0) = 2 \times 1 + 1 = 3\).

Step 3: Use the Inverse Derivative Formula

Apply the inverse derivative formula: \[ (f^{-1})'(4) = \frac{1}{f'(0)} \] Now substitute \(f'(0) = 3\) into the formula: \[ (f^{-1})'(4) = \frac{1}{3} \].

Key Concepts

Differentiation

Differentiation is a process in calculus that helps us understand how a function changes. Imagine a graph; differentiation tells us how steep the slope of the graph is at any given point. This is done by finding the derivative, which is a mathematical representation of this rate of change.
To compute a derivative, we take a function, let’s say \(f(x)\), and find its formula that gives the rate of change. The derivative is often represented as \(f'(x)\). It tells us how much \(y\) (our function value) changes with a small change in \(x\).
A derivative is crucial in understanding inverse functions, which require us to often find the derivative of the original function, as we will see in the next sections.

• It helps us calculate the rate at which things are changing.
• Derivatives can be computed using basic rules such as the power rule, product rule, and chain rule.
• Derivatives provide insights into the behavior of functions, like finding maxima, minima, or points of inflection.

Derivative of Inverse Function

The derivative of an inverse function is a bit unique because it flips the typical process of differentiation. When you have a function \(f(x)\) that is invertible, its inverse \(f^{-1}(x)\) is defined such that \(f(f^{-1}(x)) = x\).
To find the derivative of \(f^{-1}(x)\), we utilize the formula: \((f^{-1})'(b) = \frac{1}{f'(a)}\), where \(a\) is the value that, when plugged into the original function, gives \(b\) (meaning \(f(a) = b\)). This inverse derivative formula stems from the function’s symmetry in relation to the line \(y = x\).
In practical use, this means:

• Find the value of \(a\) where \(f(a) = b\).
• Compute the derivative \(f'(a)\).
• The derivative of the inverse at \(b\) is simply the reciprocal of this derivative.

It's a useful trick to calculate derivatives of complex inverses without directly finding the inverse function.

Exponential Functions

Exponential functions are a key type of function in calculus, recognizable in forms such as \(e^x\) where \(e\) is Euler’s number, approximately 2.718. These functions have unique properties that make them powerful in describing growth and decay processes
.An exponential function grows or decays at a rate proportional to its current value. This is why they often model natural processes like population growth or radioactive decay. The derivative of an exponential function like \(f(x) = e^{kx}\) is neat and straightforward: it is \(f'(x) = ke^{kx}\).

• They exhibit constant percentage growth or decay, making them predictable and particularly useful in economics and science.
• Their derivatives maintain the form of the original function, which simplifies analysis, such as when deriving inverse function derivatives.
• The function \(e^x\) has the special property that its derivative is itself, which is an essential property used in advanced calculus.

This remarkable property simplifies computations when these functions appear in calculus problems like the one in the original solution, as they show consistent behavior across transformations.