Problem 270 - Calculus Volume 1

Question

For each of the following functions, find \(\left(f^{-1}\right)^{\prime}(a).\) $$f(x)=x+\sqrt{x}, a=2$$

Step-by-Step Solution

Answer

\(\left(f^{-1}\right)^{ prime}(2) = \frac{2}{3}\)

Step 1: Understand the Problem

The problem asks for the derivative of the inverse of the function \(f(x) = x + \sqrt{x}\) evaluated at \(a = 2\). This requires us to first find the inverse function \(f^{-1}(x)\) and then compute its derivative and evaluate it at the given point.

Step 2: Find the Inverse Function

To find the inverse function \(f^{-1}(x)\), we set \(y = x + \sqrt{x}\) and solve for \(x\). Start by squaring both sides to eliminate the square root:\(y = x + \sqrt{x} \Rightarrow \sqrt{x} = y - x\). Square both sides: \(x = (y - x)^2\). Simplify and solve for \(x\) in terms of \(y\). This requires more detailed algebraic manipulation which results in: \(x = f^{-1}(y)\). For purposes of finding the derivative, we will utilize implicit differentiation rather than finding the explicit inverse.

Step 3: Apply Inverse Function Theorem

According to the inverse function theorem, the derivative of the inverse function \(\left( f^{-1} \right)'(a)\) is \(\frac{1}{f'(f^{-1}(a))}\). First, find \(x\) such that \(f(x) = a = 2\). Solve the equation: \(x + \sqrt{x} = 2\). Let \(\sqrt{x} = y\), then \(x = y^2\) and substitute: \(y^2 + y = 2\). Factoring or using the quadratic formula gives \(y = 1\) (not \(y = -2\) since \(y = \sqrt{x}\) and cannot be negative), so \(x = 1\). Thus, \(f^{-1}(2) = 1\).

Step 4: Differentiate the Original Function

Find the derivative of the original function \(f(x) = x + \sqrt{x}\). The derivative is: \(f'(x) = 1 + \frac{1}{2\sqrt{x}}\). Substitute \(x = 1\) (from \(f^{-1}(2) = 1\)) to find \(f'(1)\): \(f'(1) = 1 + \frac{1}{2\sqrt{1}} = 1 + \frac{1}{2} = \frac{3}{2}\).

Step 5: Calculate the Derivative of the Inverse Function

Using the result from Step 3 and Step 4, apply the inverse function theorem: \(\left( f^{-1} \right)'(2) = \frac{1}{f'(1)} = \frac{1}{\frac{3}{2}} = \frac{2}{3}\).

Key Concepts

Derivative of inverse function

Understanding the derivative of an inverse function can be simplified by using the Inverse Function Theorem. This theorem provides a convenient way to find the derivative of the inverse function without explicitly computing the inverse function itself. For a function \( f(x) \) that is differentiable and has an inverse function \( f^{-1}(x) \), the theorem states that the derivative of the inverse function at a point \( a \) is given by:

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

To apply this, you first need to find \( x \) such that \( f(x) = a \) and compute \( f'(x) \). Then, take the reciprocal of \( f'(x) \) at this point. This approach is efficient and avoids the complexity of directly solving for the inverse function. This method is particularly useful when dealing with functions where finding the explicit inverse is challenging, as in our example \( f(x) = x + \sqrt{x} \).

Implicit differentiation

Implicit differentiation is a powerful tool in calculus used to find derivatives of functions that are not easily expressed explicitly. It involves differentiating both sides of an equation with respect to a variable, treating other variables as functions of that variable. This method is invaluable when dealing with functions defined implicitly, such as when \( y = x + \sqrt{x} \) needs rearranging to find \( x \) in terms of \( y \).
In the given exercise, we utilized implicit differentiation to handle the complexity that arises from the presence of a square root. By rewriting \( y = x + \sqrt{x} \) and squaring both sides, you simplify the equation, making it possible to differentiate and solve for derivatives without isolating \( x \). This results in more straightforward algebraic manipulation, leading to the successful application of the Inverse Function Theorem. Implicit differentiation allows solving equations where direct differentiation would be cumbersome or impossible.

Quadratic equation solving

Solving quadratic equations is a common task in algebra, and these equations are characterized by the general form \( ax^2 + bx + c = 0 \). There are several methods to solve quadratic equations:

• Factoring: This method involves expressing the quadratic equation as a product of its linear factors.
• Quadratic formula: For any quadratic equation \( ax^2 + bx + c = 0 \), you can find the roots using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the square: This involves rewriting the quadratic in the form \( (x - p)^2 = q \), simplifying solving for \( x \).

In our exercise, solving the equation \( y^2 + y = 2 \) using quadratic techniques is pivotal. Either factoring or employing the quadratic formula gives solutions, allowing you to determine \( y \) values that correlate with \( x \) values in the context of inverse functions. In this particular problem, being able to identify the correct root, where \( y \) is non-negative, was a key step, reinforcing the relevance of quadratic skills across different mathematical areas.