Problem 62 - Thomas Calculus

Question

In Exercises \(61-66,\) you will explore some functions and their inverses together with their derivatives and linear approximating functions at specified points. Perform the following steps using your CAS: \begin{equation} \begin{array}{l}{\text { a. Plot the function } y=f(x) \text { together with its derivative over the given }} \\ {\text { interval. Explain why you know that } f \text { is one-to-one over the interval. }} \\ {\text { b. Solve the equation } y=f(x) \text { for } x \text { as a function of } y, \text { and name }} \\ {\text { the resulting inverse function } g \text { . }} \\\ {\text { c. Find the equation for the tangent line to } f \text { at the specified point }} \\ {\quad\left(x_{0}, f\left(x_{0}\right)\right) .}\\\\{\text { d. Find the equation for the tangent line to } g \text { at the point }\left(f\left(x_{0}\right), x_{0}\right)} \\ {\text { located symmetrically across the } 45^{\circ} \text { line } y=x \text { (which is the }} \\ {\text { graph of the identity function). Use Theorem } 1 \text { to find the slope }} \\ {\text { of this tangent line. }}\\\\{\text { e. Plot the functions } f \text { and } g \text { , the identity, the two tangent lines, and }} \\ {\text { the line segment joining the points }\left(x_{0}, f\left(x_{0}\right)\right) \text { and }\left(f\left(x_{0}\right), x_{0}\right) .} \\ {\text { Discuss the symmetries you see across the main diagonal. }}\end{array} \end{equation} $$y=\frac{3 x+2}{2 x-11}, \quad-2 \leq x \leq 2, \quad x_{0}=1 / 2$$

Step-by-Step Solution

Answer

The inverse is \( y = \frac{11x+2}{3-2x} \), and the function is one-to-one since it's decreasing over the interval.

Step 1: Plot the Function and Its Derivative

The function is given by \( y = \frac{3x+2}{2x-11} \). First, we compute its derivative using the quotient rule. We find that the derivative is \( f'(x) = \frac{-43}{(2x-11)^2} \). Now, plot \( y = f(x) \) and \( y = f'(x) \) over the interval \(-2 \leq x \leq 2\). The plot shows that \( f'(x) < 0 \) everywhere over the interval, indicating that \( f(x) \) is a monotonically decreasing function. Thus, it is one-to-one over this interval.

Step 2: Solve for the Inverse Function

To find the inverse function \( g \), we solve the equation \( y = \frac{3x+2}{2x-11} \) for \( x \). Rearranging gives \( x = \frac{11y+2}{3-2y} \). So, the inverse function is \( y = g(x) = \frac{11x + 2}{3 - 2x} \).

Step 3: Find the Tangent Line Equation at \( x_0 \) for \( f \)

Compute the value of \( f(\frac{1}{2}) = \frac{11}{19} \) using the original function. The derivative at \( x_0 = \frac{1}{2} \) is \( f'(\frac{1}{2}) = -\frac{43}{169} \). Therefore, the equation of the tangent line to \( f \) at this point is \[ y = -\frac{43}{169}(x - \frac{1}{2}) + \frac{11}{19}. \]

Step 4: Find the Tangent Line Equation at \( f(x_0) \) for \( g \)

The point on the inverse function \( g \) is \( (\frac{11}{19}, \frac{1}{2}) \). Use Theorem 1, which states that if \( y = f(x) \) is differentiable and \( x = g(y) \), then \( g'(y) = \frac{1}{f'(g(y))} \). Hence, \( g'(\frac{11}{19}) = -\frac{169}{43} \). The tangent line equation at this point is \[ y = -\frac{169}{43}(x - \frac{11}{19}) + \frac{1}{2}. \]

Step 5: Plot Functions and Discuss Symmetry

Plot \( y = f(x) \) and \( y = g(x) \), both tangent lines, and the identity line \( y = x \). Also, plot the line segment joining \( (\frac{1}{2}, \frac{11}{19}) \) and \( (\frac{11}{19}, \frac{1}{2}) \). Observe that across the line \( y = x \), \( f \) and \( g \) reflect each other and that their tangent lines are also reflections.

Key Concepts

Derivative

The derivative is a fundamental concept in calculus, representing the rate of change of a function with respect to its variable. For a given function \( y = \frac{3x+2}{2x-11} \), we can calculate the derivative \( f'(x) \) using the quotient rule. The quotient rule states that if you have a function \( \frac{u(x)}{v(x)} \), its derivative is given by:

• \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)

For our function, \( f'(x) \) becomes \( \frac{-43}{(2x-11)^2} \). This expression tells us how quickly and in which direction the function \( f(x) \) is changing for any given \( x \). Understanding the derivative is crucial because it helps us determine critical features of the function, such as whether it is increasing or decreasing.

Monotonically Decreasing Function

A function is said to be monotonically decreasing on an interval if it continuously decreases as the variable increases within that interval. For the function \( y = \frac{3x+2}{2x-11} \), the derivative \( f'(x) = \frac{-43}{(2x-11)^2} \) is always negative over the interval \(-2 \leq x \leq 2\).
Since \( f'(x) < 0 \) throughout this interval, \( f(x) \) is a monotonically decreasing function. This property ensures that \( f(x) \) is one-to-one, meaning each \( x \) value maps to a unique \( y \) value. A one-to-one function is key for finding an inverse function, as it means the function passes the horizontal line test and has an inverse that is also a function.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at exactly that point and has the same slope as the curve does at that point. To find the tangent line to the function \( y = f(x) \) at the point \( x_0 = \frac{1}{2} \), we use the slope \( f'(\frac{1}{2}) = -\frac{43}{169} \) and the point \( (\frac{1}{2}, \frac{11}{19}) \).
The equation of the tangent line is:

• \( y = -\frac{43}{169}(x - \frac{1}{2}) + \frac{11}{19} \)

This line gives an approximation of \( f(x) \) near \( x_0 = \frac{1}{2} \).
Tangent lines are useful for linear approximations and analyzing the behavior of functions near specific points, helping us gain insights into the function's characteristics.

Symmetry Across Lines: The Role of \( y = x \)

Symmetry across the line \( y = x \) is an important concept when dealing with inverse functions. When a function \( f(x) \) and its inverse \( g(x) \) are plotted, they reflect each other across this line. In this exercise, once we find the inverse \( y = g(x) = \frac{11x + 2}{3 - 2x} \), plotting both \( f(x) \) and \( g(x) \) alongside the identity line \( y = x \) shows this reflection.The points \( (\frac{1}{2}, \frac{11}{19}) \) and \( (\frac{11}{19}, \frac{1}{2}) \) also demonstrate symmetry because they lie exactly across from each other over the line \( y = x \).
The symmetry principle tells us that the tangent to \( f \) at a point will have a counterpart tangent to \( g \) at the mirrored point. This consistent relationship provides a visual and conceptual insight into how inverse functions relate through symmetry across the line \( y = x \).