Problem 66 - 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} $$y=2-x-x^{3}, \quad-2 \leq x \leq 2, \quad x_{0}=\frac{3}{2}$$

Step-by-Step Solution

Answer

The function is one-to-one as it is strictly decreasing. The inverse satisfies the cubic equation. Tangents show symmetrical reflection across the identity line.

Step 1: Plot the Function and its Derivative

Begin by plotting the function \( y = f(x) = 2-x-x^3 \) over the interval \(-2 \leq x \leq 2\). Compute its derivative: \( f'(x) = -1 - 3x^2 \). On analyzing \( f'(x) \), note that it is negative for all \( x \) in the interval, implying the function is strictly decreasing, which means it is one-to-one.

Step 2: Solve for the Inverse Function

To find the inverse, set \( y = 2 - x - x^3 \) and solve for \( x \) in terms of \( y \). Rearranging gives a cubic equation: \( x^3 + x + y - 2 = 0 \). Use computational tools to find that the inverse function \( g(y) \) satisfies this cubic equation.

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

At \( x_0 = \frac{3}{2} \), calculate \( f(x_0) = 2 - \frac{3}{2} - \left(\frac{3}{2}\right)^3 = -\frac{11}{8} \). The slope of the tangent line is \( f'(x_0) = -1 - 3\left(\frac{3}{2}\right)^2 = -\frac{29}{4} \). The equation of the tangent line at \( x_0 \) is \( y + \frac{11}{8} = -\frac{29}{4}(x - \frac{3}{2}) \) or \( y = -\frac{29}{4}x + \frac{79}{16} \).

Step 4: Find the Tangent Line to g

The point \((f(x_0), x_0) = (-\frac{11}{8}, \frac{3}{2})\) lies on the graph of \( g \). The slope of the tangent line to \( g \) at this point is the reciprocal of the slope of \( f \) at \( x_0 \), which is \(-\frac{4}{29}\). The equation of this tangent line is \( x - \frac{3}{2} = -\frac{4}{29} \left(y + \frac{11}{8}\right) \) or \( x = -\frac{4}{29}y + \frac{37}{16} \).

Step 5: Plot and Discuss Symmetries

Plot the functions \( f \) and \( g \), the identity line \( y = x \), both tangent lines, and the line segment joining points \( (\frac{3}{2}, -\frac{11}{8}) \) and \( (-\frac{11}{8}, \frac{3}{2}) \). Notice that the reflection about the line \( y=x \) is evident in the symmetry between \( f \) and \( g \) and their respective tangent lines about the 45° line. This illustrates how each point on \( f \) and \( g \) are mirrors of each other across the identity line.

Key Concepts

Derivatives

Understanding derivatives is crucial for analyzing the behavior of functions. A derivative represents the rate of change of a function's output with respect to changes in its input. In the context of finding inverse functions, derivatives help us verify if a function is one-to-one.

A negative derivative across an interval, as seen in \( f'(x) = -1 - 3x^2 \), indicates that the function is strictly decreasing. This means for every increase in \( x \), \( y \) decreases, ensuring no repeated \( y \) values, thus, the function is one-to-one.

• Derivatives can be used to determine points of interest, like maxima or minima.
• They provide essential information for curve sketching.
• Inverses require the function to be one-to-one, which can be ascertained through derivative analysis.

One-to-One Functions

One-to-one functions are fundamental in creating inverse functions. A function \( f(x) \) is considered one-to-one if each output is only produced by a single input, meaning the function either increases or decreases throughout its domain.

In our example with \( y = 2-x-x^3 \), the derivative confirmed that the function is strictly decreasing on \(-2 \leq x \leq 2\). This means \( f(x) \) passes the Horizontal Line Test: any horizontal line crosses the function \( f(x) \) at most once.

• One-to-one functions are critical for inverse functions since inverses reverse the role of inputs and outputs.
• To validate a function’s one-to-one nature, it must consistently increase or decrease.

Understanding this concept is key to comprehending how to find and use inverses accurately.

Tangent Lines

Tangent lines are straight lines that touch a curve at a single point. They share the same slope as the curve does at that point, providing a linear approximation of the curve near this point.

To find the tangent line of a function \( f(x) \) at a given point, you need the slope of the curve, provided by its derivative evaluated at that point, and the point itself. For example, at \( x_0 = \frac{3}{2} \), the slope of the tangent line to \( f(x) \) is \(-\frac{29}{4}\), leading us to the equation \( y = -\frac{29}{4}x + \frac{79}{16} \).

Tangent lines are invaluable for:

• Estimating values of functions at points close to the tangent point.
• Analyzing changes in functions, as they capture the direction and rate of change.

They also play an important role in understanding symmetrical properties when studying inverse functions.

Function Plotting

Plotting functions and their various elements, such as derivatives or inverse functions, helps provide a visual understanding of mathematical concepts. Visual graphing allows us to examine symmetry, intersection points, and behavior of functions over intervals.

In this exercise, plotting \( y = f(x) = 2-x-x^3 \), its derivative, and its inverse \( g(y) \), illustrates how these functions relate to each other and the identity line \( y = x \).

Function plotting is useful for:

• Visualizing the differences and similarities between \( f(x) \) and \( g(y) \).
• Highlighting symmetries such as mirroring across the line \( y=x \).
• Understanding the geometrical significance of mathematical operations.

Overall, graphing provides a comprehensive insight that numerical calculations or symbolic computations alone may not offer.