Problem 61 - 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} $$y=\sqrt{3 x-2}, \frac{2}{3} \leq x \leq 4, \quad x_{0}=3$$

Step-by-Step Solution

Answer

The function \( y=\sqrt{3x-2} \) is one-to-one. Its inverse is \( g(y)=\frac{y^2+2}{3} \). The tangent lines demonstrate reflection symmetry across the line \( y=x \).

Step 1: Plot Function and Derivative

Firstly, let's plot the function \( y=\sqrt{3x-2} \) and its derivative over the interval \( \frac{2}{3} \leq x \leq 4 \). The derivative is given by \( f'(x) = \frac{3}{2\sqrt{3x-2}} \). By plotting the derivative, we see that it is always positive within the interval, which means the function is strictly increasing and hence one-to-one.

Step 2: Solve for Inverse Function

To find the inverse, solve the equation \( y=\sqrt{3x-2} \) for \( x \). Squaring both sides gives \( y^2 = 3x - 2 \). Solving for \( x \) will yield \( x = \frac{y^2 + 2}{3} \). The inverse function is \( g(y) = \frac{y^2 + 2}{3} \).

Step 3: Find Tangent Line to Function

The point of tangency on the function is \( (3, f(3)) \). Compute \( f(3) = \sqrt{3(3)-2} = \sqrt{7} \). The slope of the tangent line is \( f'(3) = \frac{3}{2\sqrt{3(3)-2}} = \frac{3}{2\sqrt{7}} \). The equation of the tangent line is \( y - \sqrt{7} = \frac{3}{2\sqrt{7}}(x - 3) \).

Step 4: Find Tangent Line to Inverse Function

For the inverse, the tangent point is \( (\sqrt{7}, 3) \). Use Theorem 1, which relates the slope of the inverse tangent to the reciprocal of the original function's tangent slope at a swapped point. The slope is \( \frac{2\sqrt{7}}{3} \). The equation is \( y - 3 = \frac{2\sqrt{7}}{3}(x - \sqrt{7}) \).

Step 5: Plot Functions and Tangent Lines

On a single graph, plot the function \( f(x) \), its inverse \( g(y) \), the identity line \( y=x \), the two tangent lines found in Steps 3 and 4, and the line segment joining the points \((3,\sqrt{7})\) and \((\sqrt{7},3)\). Notice the symmetric relationship of the tangents about \( y = x \). This shows the reflectional symmetry of a function and its inverse on this line.

Key Concepts

One-to-One Functions

A function is considered one-to-one when each input maps to a unique output. In other words, no two different inputs yield the same output. This property is crucial for a function to have an inverse. An easy way to recognize a one-to-one function is by using the horizontal line test. If no horizontal line intersects the graph of the function more than once, then the function is one-to-one.

For our function, \( y = \sqrt{3x-2} \), examining its derivative, \( f'(x) = \frac{3}{2\sqrt{3x-2}} \), provides clarity. Since this derivative is positive throughout the interval \( \frac{2}{3} \leq x \leq 4 \), the function is strictly increasing, guaranteeing that it is one-to-one. As such, this function will have an inverse function mapping outputs back to their unique inputs.

Derivative of a Function

The derivative of a function gives us information about how the function's output changes in response to changes in the input. It measures the slope of the tangent line to the graph at any point. For the function \( y = \sqrt{3x-2} \), the derivative is calculated as \( f'(x) = \frac{3}{2\sqrt{3x-2}} \).

This derivative tells us how sharply or gradually the function rises or falls at any given point. Since our derivative is always positive within the interval \( \frac{2}{3} \leq x \leq 4 \), the function is increasing. Thus, knowing the derivative not only indicates the behavior of the function over an interval but also helps in finding the equation of the tangent line. The tangent is crucial when discussing properties like smoothness and continuity of functions.

Tangent Line

The tangent line to a function at a given point is a straight line that just "touches" the function, having the same slope as the function at that point. The equation of the tangent line is used to approximate the function near the point of tangency. In our case, we find the tangent to \( y = \sqrt{3x-2} \) at \( x_0 = 3 \).

We first calculate the slope of the tangent line using the derivative: \( f'(3) = \frac{3}{2\sqrt{7}} \). We know that the function's value at this point is \( \sqrt{7} \). The tangent line can then be described by the equation \( y - \sqrt{7} = \frac{3}{2\sqrt{7}}(x - 3) \). This line is particularly useful in calculating linear approximations of the function near the specified point, illustrating the concept of local linearity.

Reflection Symmetry

Reflection symmetry in mathematics refers to a scenario where one half of an object or design is a mirror image of the other half. In the context of functions and their inverses, this type of symmetry becomes apparent when both are plotted on the same graph alongside the line \( y = x \).

For the function \( y = \sqrt{3x-2} \) and its inverse \( g(y) = \frac{y^2 + 2}{3} \), plotting these alongside the identity line \( y = x \) reveals a symmetry. The points of intersection across this line establish this reflection symmetry. This means that each point \( (x,y) \) on the function corresponds to a point \( (y,x) \) on its inverse, creating a symmetric reflection about the line \( y = x \). This kind of symmetry not only helps in visualizing the relationships between functions and their inverses but also in understanding how tangent lines at points swap roles between the function and its inverse.