Problem 7 - 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. $$ f(\sqrt{3})=4, f^{\prime}(s)=\left(2+s^{2}\right) /\left(1+s^{2}\right) $$

Step-by-Step Solution

Answer

The derivative of the inverse at 4 is \(\frac{4}{5}\).

Step 1: Understanding the Inverse Function Formula

The derivative of an inverse function \((f^{-1})'\) is given by the formula \((f^{-1})'(y) = \frac{1}{f'(x)}\), where \(f(x) = y\). This means we need to identify the \(x\) such that \(f(x) = 4\), and use \(f'(x)\) in our calculation.

Step 2: Identify the Necessary x-Value

Since \(f(\sqrt{3}) = 4\), we know \(x = \sqrt{3}\) corresponds to \(y = 4\). Thus, the derivative of the inverse function at \(y = 4\) will depend on \(f'(\sqrt{3})\).

Step 3: Compute f'(√3)

Use the given function \(f'(s) = \frac{2+s^2}{1+s^2}\). Substitute \(s = \sqrt{3}\): \[ f'(\sqrt{3}) = \frac{2+3}{1+3} = \frac{5}{4}. \]

Step 4: Calculate the Derivative of the Inverse Function

Using the inverse function formula, \((f^{-1})'(4) = \frac{1}{f'(\sqrt{3})}\). Thus, \[ (f^{-1})'(4) = \frac{1}{\frac{5}{4}} = \frac{4}{5}. \]

Key Concepts

Derivative

The concept of a derivative is central to calculus and is all about understanding how functions change. When we find the derivative of a function, we're looking at the function's rate of change at any point.

A good way to imagine this is by considering the slope of a tangent line to a curve at a point. This slope tells us how steep the curve is at that precise spot. Mathematically, if we have a function \( f(x) \), its derivative is often written as \( f'(x) \) or \( \frac{df}{dx} \).

The derivative can provide crucial information about the function. This includes where the function is increasing or decreasing, where it might have a maximum or minimum, and where it is flat or changes direction. In the context of this exercise, knowing the derivative helps us understand how outputs of a function change with respect to inputs.

Invertible Function

An invertible function is one that has a unique output for every unique input. This means that it can be "reversed," so to speak. In other words, for a function to be invertible, there must be an inverse function that can take the outputs of the original function back to their corresponding inputs.

Mathematically, for a function \( f \) to be invertible, each element in its range must be mapped from exactly one element in its domain. We say that such a function is one-to-one and onto. This one-to-one relationship is crucial for the function's inverse to exist.

• Example: If \( f(x) = y \), then there must be an \( f^{-1}(y) = x \).
• Conditions for invertibility often involve the function being continuous and strictly monotonic (either entirely non-increasing or non-decreasing).

Recognizing whether a function is invertible is key to solving problems involving inverse functions.

Inverse Function Theorem

The inverse function theorem is a powerful tool in calculus that helps us determine when a function has an inverse and how to find the derivative of that inverse. This theorem states that if a function \( f \) is continuously differentiable and its derivative \( f'(x) \) is not zero at a point, then \( f \) has an inverse in a neighborhood around that point.

More importantly, for the exercise at hand, the inverse function theorem gives us a way to calculate the derivative of the inverse function. Specifically, it tells us that \((f^{-1})'(y) = \frac{1}{f'(x)}\), where \( f(x) = y \).

• This formula is essential when you have the original function's derivative and need the inverse function's derivative.
• Applying it requires identifying \( x \) for which \( f(x) = y \). Once \( x \) is known, you can find \( f'(x) \) and then \((f^{-1})'(y) \).

This theorem simplifies handling inverse function problems, especially when dealing with differentiation and continuity.