Problem 36 - Thomas Calculus

Question

In Exercises \(35-38 :\) \begin{equation} \begin{array}{l}{\text { a. Find } f^{-1}(x) \text { . }} \\ {\text { b. Graph } f \text { and } f^{-1} \text { together. }} \\ {\text { c. Evaluate } d f / d x \text { at } x=a \text { and } d f^{-1} / d x \text { at } x=f(a) \text { to show }} \\ {\text { that at these points } d f^{-1} / d x=1 /(d f / d x)}\end{array} \end{equation} $$f(x)=\frac{x+2}{1-x}, \quad a=1 / 2$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{x-2}{x+1} \), and the derivatives confirm their reciprocal relationship.

Step 1: Find the Inverse Function

To find the inverse function \( f^{-1}(x) \), start by setting \( y = f(x) = \frac{x+2}{1-x} \). Then switch \( x \) and \( y \), resulting in \( x = \frac{y+2}{1-y} \). Solve for \( y \) by cross-multiplying to get \( x(1-y) = y+2 \). This simplifies to \( y = \frac{x-2}{x+1} \). Therefore, the inverse function is \( f^{-1}(x) = \frac{x-2}{x+1} \).

Step 2: Graph the Functions

To graph \( f(x) \) and \( f^{-1}(x) \), plot both functions on the same coordinate system. For \( f(x) = \frac{x+2}{1-x} \), identify important points such as the y-intercept, asymptotes, and behavior at infinity. Similarly, for \( f^{-1}(x) = \frac{x-2}{x+1} \), identify the same characteristics. Include the line \( y = x \), since the graphs of a function and its inverse are reflections across this line.

Step 3: Differentiate the Function

Differentiate \( f(x) = \frac{x+2}{1-x} \) with respect to \( x \) to find \( \frac{df}{dx} \). Using the quotient rule, set \( u = x+2 \) and \( v = 1-x \). Then \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \). This gives \( \frac{df}{dx} = \frac{(1)(1-x) - (x+2)(-1)}{(1-x)^2} = \frac{3}{(1-x)^2} \).

Step 4: Differentiate the Inverse Function

Differentiate \( f^{-1}(x) = \frac{x-2}{x+1} \) with respect to \( x \) using the quotient rule again. Let \( u = x-2 \) and \( v = x+1 \), giving \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 1 \). The derivative is \( \frac{df^{-1}}{dx} = \frac{(1)(x+1) - (x-2)(1)}{(x+1)^2} = \frac{3}{(x+1)^2} \).

Step 5: Evaluate Derivatives at Specific Points

Evaluate \( \frac{df}{dx} \) at \( x = \frac{1}{2} \): \( \frac{df}{dx} \bigg|_{x = \frac{1}{2}} = \frac{3}{\left(1-\frac{1}{2}\right)^2} = \frac{3}{\frac{1}{4}} = 12 \). Now evaluate \( \frac{df^{-1}}{dx} \) at \( x = f\left(\frac{1}{2}\right) = 5 \): \( \frac{df^{-1}}{dx} \bigg|_{x = 5} = \frac{3}{(5+1)^2} = \frac{3}{36} = \frac{1}{12} \).

Step 6: Confirm the Relationship of Derivatives

The relationship \( \frac{df^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \) holds true at the corresponding points, as shown before: \( \frac{df}{dx} = 12 \) and \( \frac{df^{-1}}{dx} = \frac{1}{12} \). These are indeed reciprocals, confirming that \( \frac{df^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus representing the rate of change of a function. They allow us to determine how a function's output changes as its input changes slightly. In this exercise, differentiating the function \( f(x) = \frac{x+2}{1-x}\) helped us understand how \( f(x)\) behaves as \( x\) changes.

To find the derivative \( \frac{df}{dx} \), we employ the quotient rule. This rule is particularly useful when the function is expressed as a ratio of two functions, like \( \frac{u}{v}\), where \( u = x+2\) and \( v = 1-x\).

The quotient rule states that:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

In our case, it simplifies to:

\[ \frac{df}{dx} = \frac{(1)(1-x) - (x+2)(-1)}{(1-x)^2} = \frac{3}{(1-x)^2} \]

This derivative tells us how \( f(x)\) changes at any point \( x\). By computing this, we can then evaluate the specific rate of change for any given \( x \) value.

Quotient Rule

The quotient rule is an essential technique in calculus for finding derivatives of functions that are ratios of two other functions. It's a key process when working with inverse functions, as seen in our exercise.

When you have a function \( f(x) = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions, the quotient rule lets you differentiate \( f(x)\) efficiently.

• Let \( u = x+2 \) and \( v = 1-x \).
• The derivatives are \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \).
• Apply the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{(v \cdot \frac{du}{dx}) - (u \cdot \frac{dv}{dx})}{v^2} \).
• This gives \( \frac{3}{(1-x)^2} \).

For inverse functions, such as \( f^{-1}(x) = \frac{x-2}{x+1} \) in this exercise, apply the quotient rule again:

Let \( u = x-2 \) and \( v = x+1 \), resulting in \( \frac{df^{-1}}{dx} = \frac{3}{(x+1)^2} \).

This method ensures you compute derivatives accurately and is vital for verifying how functions and their inverses interact.

Graphing Functions

Graphing functions and their inverses offers a visual representation of the relationship between them. By plotting both \( f(x) \) and its inverse \( f^{-1}(x) \), you can observe key properties like reflection and asymptotic behavior.

The process involves a few steps:

• Identify characteristics such as y-intercepts and asymptotes of the functions.
• Plot several points for each function to understand their shape.
• Include the line \( y = x \) on your graph. This acts as a mirror line, as \( f(x) \) and \( f^{-1}(x) \) are reflections of each other across this line.

For our functions, \( f(x) = \frac{x+2}{1-x} \) and \( f^{-1}(x) = \frac{x-2}{x+1} \), graphing highlights their symmetrical nature relative to the line \( y = x \). Observing the graphs helps reinforce the concept that finding an inverse function essentially means "flipping" a function over the line \( y = x \).

This visualization not only aids in understanding the reciprocal relationship of derivatives but also solidifies the concept of what an inverse function represents.

Reciprocal Relationship

The reciprocal relationship between a function and its inverse is elegantly showcased in their derivatives. This relationship means that the derivative of the inverse is the reciprocal of the original function's derivative at corresponding points.

For the function \( f(x) = \frac{x+2}{1-x} \), we found \( \frac{df}{dx} = \frac{3}{(1-x)^2} \). At \( x = \frac{1}{2} \), \( \frac{df}{dx} \bigg|_{x = \frac{1}{2}} = 12 \).

For the inverse function \( f^{-1}(x) = \frac{x-2}{x+1} \), evaluated at \( f\left(\frac{1}{2}\right) = 5 \), we have: \( \frac{df^{-1}}{dx} \bigg|_{x = 5} = \frac{1}{12} \). The derivatives are:

• \( \frac{df}{dx} = 12 \)
• \( \frac{df^{-1}}{dx} = \frac{1}{12} \)

These results are reciprocals of one another, confirming the reciprocal relationship property where \( \frac{df^{-1}}{dx} = \frac{1}{\left(\frac{df}{dx}\right)} \).

This principle is crucial when verifying solutions for inverse functions as it ties together the changes in the original function and its inverse, reflecting broader mathematical interconnectedness.