Problem 64 - 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} $$y=\frac{x^{3}}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$

Step-by-Step Solution

Answer

The function is one-to-one over the interval; its inverse is \( g(y) \). Tangent lines reflect symmetry across \( y = x \).

Step 1: Plot the Function and Derivative

First, we plot the function \( y = \frac{x^3}{x^2 + 1} \) and its derivative over the interval \(-1 \le x \le 1\). The derivative is calculated as \( f'(x) = \frac{3x^2(x^2 + 1) - 2x^4}{(x^2 + 1)^2} \). Since the derivative \( f'(x) \) is strictly positive or negative (and not zero) over the interval, \( f \) is one-to-one over \([-1, 1]\).

Step 2: Solve for the Inverse Function

To find the inverse, solve the equation \( y = \frac{x^3}{x^2 + 1} \) for \( x \) in terms of \( y \). After rearranging and solving, the inverse function is named as \( g(y) = x \). In practice, this requires using a computer algebra system (CAS) due to complexity.

Step 3: Find Tangent Line to \( f \)

Calculate the equation of the tangent line to \( f \) at the point \( x_0 = \frac{1}{2} \). The slope of this tangent line is \( f'\left(\frac{1}{2}\right) \), and the tangent line is \( y - f\left(\frac{1}{2}\right) = f'\left(\frac{1}{2}\right)(x - \frac{1}{2}) \). Plug in values to obtain the specific equation.

Step 4: Find Tangent Line to Inverse \( g \)

Using Theorem 1, the slope of the tangent line to \( g \) at \( (f(x_0), x_0) \) is the reciprocal of the slope at \( (x_0, f(x_0)) \). The equation of this tangent line is \( x - x_0 = \frac{1}{f'(x_0)}(y - f(x_0)) \). Substitute the known values to find this equation.

Step 5: Plot and Discuss Symmetries

Plot the functions \( f(x) \) and \( g(x) \), the identity line \( y = x \), the tangent lines found in Steps 3 and 4, and the line segment connecting \( (x_0, f(x_0)) \) and \( (f(x_0), x_0) \). Observe and discuss the symmetry across the line \( y = x \), showing that \( g \) is a reflection of \( f \) over this diagonal.

Key Concepts

Derivative of a function

The derivative of a function provides critical information about the rate of change of the function's output with respect to changes in the input. This concept is fundamental in calculus. When you differentiate a function, you are essentially finding the slope of the tangent to the function at any point.
This slope is what informs us about the function's increasing or decreasing behavior over specific intervals.
- A positive derivative indicates that the function is increasing. - A negative derivative shows that the function is decreasing. - If the derivative is zero, it's possible that there's a local maximum or minimum at that point.
For inverse functions, the derivative plays a key role in understanding the one-to-one nature of functions. In other words, a strict positive or negative derivative over an interval suggests that the function has an inverse, because it implies that the function is monotonically increasing or decreasing.

Linear approximation

Linear approximation is a technique used to estimate the value of a function near a point using the tangent line of the function at that point. This is very useful when you want to understand how a function behaves locally without needing complex calculations.
The formula for linear approximation is derived from the equation of a tangent line. For a function \( f(x) \) at a point \( x = a \), the linear approximation is given by:\[L(x) = f(a) + f'(a) \times (x - a)\]
This linear function, \( L(x) \), is easy to compute and provides an approximate solution for \( f(x) \) close to \( a \).
- It's most accurate near \( a \) because the tangent line closely follows the curve at this point.- The further the value of \( x \) from \( a \), the less accurate the approximation.
This concept is directly applied when calculating the tangent line of functions, giving a valid reason for understanding and working with slopes in function analysis.

Tangent line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of this line is equal to the derivative of the function at that point.
Finding the tangent line involves a few important steps:

• Identify the point of tangency
• Calculate the slope at this point using the derivative
• Formulate the equation of the line using point-slope form: \( y - y_0 = m(x - x_0) \), where \( m \) is the slope and \( (x_0, y_0) \) are the coordinates of the point of tangency.

The tangent line serves many purposes:- It provides a linear approximation.- It indicates the function's behavior at that specific point.- It helps in analyzing the inverse functions and their behavior geometrically when reflecting over the identity \( y = x \) line, observing symmetry.

Symmetry of functions

Symmetry in functions can refer to several types of visual and analytical properties observed in graphs. One common type of symmetry is reflectional symmetry, where one part of a graph mirrors another across a certain line.
In inverse functions, symmetry often implies reflection over the line \( y = x \). When a function \( f \) and its inverse \( g \) are plotted, they reflect across the identity line, showcasing their beautifully symmetric relationship.
Some key points about symmetry in functions are:

• For a pair of functions and their inverses, each point \( (a, b) \) on \( f \) corresponds to \( (b, a) \) on \( g \).
• This symmetry is especially evident when considering their tangent lines, as one tangent at \( f \)'s point of tangency will mirror \( g \)'s tangent at the corresponding symmetric point.
• Understanding symmetry helps in graph analysis and simplifies the visual interpretation of the inverse relationship.

Symmetry doesn't only provide aesthetic appeal but also reinforces the mathematical properties connecting a function and its inverse.