Problem 271 - Calculus Volume 1

Question

For each of the following functions, find \(\left(f^{-1}\right)^{\prime}(a).\) $$f(x)=x-\frac{2}{x}, x<0, a=1$$

Step-by-Step Solution

Answer

The derivative of the inverse function at \( a = 1 \) is \( \frac{1}{3} \).

Step 1: Recall the Relationship for Derivative of Inverse Function

To find the derivative of the inverse function at a given point, we use the formula: \ \( \left(f^{-1}\right)^{\prime}(a) = \frac{1}{f^{\prime}(f^{-1}(a))} \). \ Our task is to find \( f^{-1}(a) \) first and then calculate \( f^{\prime}(x) \).

Step 2: Set f(x) Equal to a

For \( a = 1 \), we need to solve \( f(x) = 1 \). \ This means solving \( x - \frac{2}{x} = 1 \) for \( x \).

Step 3: Solve the Equation x - \frac{2}{x} = 1

Multiply by \( x \) to clear the fraction: \( x^2 - 2 = x \). \ Rearrange to: \( x^2 - x - 2 = 0 \). \ Factor the quadratic: \( (x - 2)(x + 1) = 0 \). \ Since \( x < 0 \), choose \( x = -1 \). \ Therefore, \( f^{-1}(1) = -1 \).

Step 4: Find the Derivative f'(x)

Calculate \( f^{\prime}(x) \) by differentiating \( f(x) = x - \frac{2}{x} \): \ \( f^{\prime}(x) = 1 + \frac{2}{x^2} \).

Step 5: Evaluate f'(x) at x = f^{-1}(1)

Plug \( x = -1 \) into \( f^{\prime}(x) = 1 + \frac{2}{x^2} \): \ \( f^{\prime}(-1) = 1 + \frac{2}{(-1)^2} = 1 + 2 = 3 \).

Step 6: Compute the Derivative of the Inverse Function

Using the formula from Step 1: \ \( \left(f^{-1}\right)^{\prime}(1) = \frac{1}{f^{\prime}(-1)} = \frac{1}{3} \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point.
When learning about inverse functions, understanding the derivative is crucial. This is because the derivative of an inverse function helps us find the slope of the original function at a corresponding point.

• To calculate the derivative of a function, we look at the limit of the function as the change in the input variable becomes infinitesimally small. This is often represented by the formula, \( f^{ ext{'}(x)} = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} \).
• For inverse functions, the relationship for the derivative is given by the formula: \( \left(f^{-1}\right)^{ ext{'}(a)} = \frac{1}{f^{ ext{'}}(f^{-1}(a))} \).
• This formula shows that to find the derivative of the inverse function at a point, we first need the derivative of the original function at the corresponding point of the inverse.

In our example, after identifying the inverse value, we find the derivative of the original function at this value before applying the inverse derivative formula.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, generally expressed in the form \( ax^2 + bx + c = 0 \) where \( a, b, \) and \( c \) are constants.
Solving quadratic equations can be done through a variety of methods like factoring, using the quadratic formula, or completing the square.

• In the context of finding inverse functions, they often appear when rearranging the expressions to solve for a variable.
• When given an equation like \( x - \frac{2}{x} = 1 \), you clear the fraction by multiplying through by \( x \), leading to \( x^2 - x - 2 = 0 \).
• The next step is to factor this quadratic equation to find its roots. In this example, it factors to \( (x - 2)(x + 1) = 0 \), giving roots of \( x = 2 \) and \( x = -1 \).
• When working with quadratic equations, it's critical to consider the domain of the function. Since we specified \( x < 0 \) in this exercise, we choose \( x = -1 \) as our solution.

Understanding how to work with quadratic equations is essential in solving for inverses or simplifying problem steps when working with functions.

Function Composition

Function composition involves applying one function to the results of another. It's like layering functions over each other, where the output of one becomes the input for another. This technique is vital when dealing with invertible functions and their derivatives.

• The notation for function composition is \( (f \circ g)(x) = f(g(x)) \), meaning we substitute the result of \( g(x) \) into the function \( f \).
• When looking at inverse functions, composition helps verify if functions truly are inverses. If \( f \) and \( g \) are inverses, then \( f(g(x)) = x \) and \( g(f(x)) = x \).
• In the context of finding derivatives of inverse functions, understanding composition assists in understanding how changes in one function affect another.
• Through function composition, we analyze how a function maps to its inverse and determine the impact of such mappings in terms of slopes and changes, as seen in the derivative formulas.

Mastering function composition allows us to approach inverse functions logic, significantly easing the understanding of their derivatives.