Problem 271 - Calculus Volume 1

Question

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: Understanding Inverse Derivative Formula

To find the derivative of the inverse function at a point, we use the formula \(rac{1}{f'(f^{-1}(a))}\). We need to find \(f^{-1}(a)\) first.

Step 2: Solve for Inverse Function Value

Since we know \(f(x) = x - \frac{2}{x}\), we need \(f(x) = 1\) to find \(f^{-1}(1)\). Setting \(x - \frac{2}{x} = 1\) gives \(x^2 - x - 2 = 0\). Solve this quadratic equation next.

Step 3: Solve the Quadratic Equation

The quadratic equation \(x^2 - x - 2 = 0\) can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -1\), and \(c = -2\). Calculate the discriminant \(\Delta = (-1)^2 - 4(1)(-2) = 1 + 8 = 9\).

Step 4: Calculate Roots of the Quadratic

The roots are \(x = \frac{1 \pm 3}{2}\). This gives \(x = 2\) or \(x = -1\). Since \(x < 0\), we take \(x = -1\). Therefore, \(f^{-1}(1) = -1\).

Step 5: Find the Derivative of \(f(x)\)

Differentiate \(f(x) = x - \frac{2}{x}\). The derivative \(f'(x) = 1 + \frac{2}{x^2}\).

Step 6: Evaluate Derivative at Inverse Value

Substitute \(x = -1\) into \(f'(x) = 1 + \frac{2}{x^2}\). This gives \(f'(-1) = 1 + \frac{2}{1} = 3\).

Step 7: Calculate Reciprocal for \((f^{-1})'(a)\)

Use the inverse function derivative formula \((f^{-1})'(1) = \frac{1}{f'(-1)} = \frac{1}{3}\).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). Solving these equations is key in various areas of mathematics and engineering. They can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps find the roots of the quadratic, which are the values of \( x \) that satisfy the equation.To apply the formula, calculate the discriminant \(\Delta = b^2 - 4ac\). It tells you how many real roots the equation has:- If \(\Delta > 0\), there are two distinct real roots.- If \(\Delta = 0\), there is one real root.- If \(\Delta < 0\), the roots are complex.
In the exercise, the quadratic equation \(x^2 - x - 2 = 0\) is solved using these steps, and the roots calculated give values that are essential for determining the inverse function value.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any point on its curve.

• The derivative is often denoted as \(f'(x)\) or \(\frac{dy}{dx}\) for functions \(y = f(x)\).

In the given exercise, differentiation is applied to the function \(f(x) = x - \frac{2}{x}\) to find its derivative:

• First, identify the derivative \(f'(x) = 1 + \frac{2}{x^2}\).

This step is crucial because finding the derivative allows us to calculate the inverse derivative later on.
Differentiation is not just used for finding slopes; it is also widely used in physics for velocity, acceleration, and more.

Inverse Derivative Formula

The inverse derivative formula helps us find the derivative of an inverse function at a particular point.It is given by:

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

This formula is essential when dealing with inverse functions, enabling us to find the rate of change of the original variable relative to the inverse function.For the exercise provided, once we found \(f^{-1}(1) = -1\), we plugged this value into the derivative of the original function \(f'(x)\) to find \(f'(-1) = 3\).Finally, applying the inverse derivative formula, we have:

• \((f^{-1})'(1) = \frac{1}{3}\).

This calculation shows how the rate of change of the inverse function at \(a = 1\) is simply the reciprocal of the derivative of the function at \(x = -1\), illustrating the beautiful interplay between differentiation and inverse functions.