Problem 60 - Thomas Calculus

Question

In Exercises \(53-60,\) 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: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\) . c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right) .\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\) , the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal. $$ y=\sin x, \quad-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}, \quad x_{0}=1 $$

Step-by-Step Solution

Answer

Functions \(f\) and \(g\) are mirrored across \(y = x\) with tangent lines reciprocal at given points.

Step 1: Plot and Analyze Function and Derivative

First, plot the function \( y = \sin x \) over the interval \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). Its derivative is \( y' = \cos x \). The cosine function is positive over this interval, which implies that \( \sin x \) is strictly increasing and thus one-to-one.

Step 2: Find Inverse Function

To find the inverse, solve \( y = \sin x \) for \( x \). This gives \( x = \arcsin y \). Hence, the inverse function is \( g(y) = \arcsin y \).

Step 3: Find Tangent Line to f at Specified Point

The point of tangency is \( (1, \sin 1) \). The derivative \( y' = \cos x \) evaluated at \( x = 1 \) is the slope of the tangent line. So the slope is \( \cos 1 \). The tangent line's equation is \( y - \sin 1 = \cos 1 (x - 1) \).

Step 4: Find Tangent Line to Inverse Function at Point

The corresponding point for \( g \) is \( (\sin 1, 1) \). By Theorem 1, the slope of the tangent to \( g \) is the reciprocal of the slope of the tangent to \( f \), meaning \( \frac{1}{\cos 1} \). Therefore, the equation is \( y - 1 = \frac{1}{\cos 1}(x - \sin 1) \).

Step 5: Plot Functions and Tangent Lines

Plot \( y = \sin x \), \( y = \arcsin y \), the identity line \( y = x \), both tangent lines, and the line segment connecting \( (1, \sin 1) \) to \( (\sin 1, 1) \). Observe that the plots of \( f \) and \( g \) are mirrored across the line \( y = x \), reflecting the inverse relationship.

Key Concepts

Derivatives

In calculus, the derivative of a function at any point gives us the rate at which the function's value changes at that point. It is essentially the slope of the tangent line to the graph of the function at a given point. For example, in the exercise where the function is \( y = \sin x \), the derivative is \( y' = \cos x \). This derivative is crucial because it helps us understand how \( y = \sin x \) behaves over its interval.
Over the interval \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\), \(\cos x\) remains positive, indicating that the function \(\sin x\) is increasing consistently.
This consistent increase signifies that the function doesn't return the same y-values for different x-values, meaning \(\sin x\) is one-to-one over the given interval. A strictly positive (or negative) derivative over an interval ensures that a function is one-to-one on that interval.

Tangent Lines

A tangent line to a curve at a given point is a straight line that just touches the curve at that point and has the same slope as the curve at that point. It provides a linear approximation of the curve's behavior near the point of tangency.
For the function \( y = \sin x \) at \( x = 1 \), the slope is given by its derivative, \( \cos 1 \). Thus, the equation of the tangent line is \( y - \sin 1 = \cos 1 (x - 1) \).
This line gives a linear approximation of the sine curve near the point \( (1, \sin 1) \). When considering the inverse function \( g(y) = \arcsin y \), the slope at \( (\sin 1, 1) \) is crucial for understanding symmetry, calculated as \( \frac{1}{\cos 1} \). This reciprocal relationship highlights the reverse nature of functions and their inverses geometrically.
Understanding tangent lines is fundamental for analyzing the linear properties of a non-linear function at a specific point.

One-to-One Functions

A function is described as one-to-one (or injective) if it never assigns the same y-value to different x-values. This property is visually identifiable in a graph where any horizontal line intersects the graph at most once.
The exercise demonstrates the function \( y = \sin x \) within the restricted interval \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \), where the function is strictly increasing due to a positive derivative \( y' = \cos x \). Such strict monotonicity (always increasing or decreasing) guarantees that \( y = \sin x \) is one-to-one in this context.
The one-to-one nature is pivotal in calculus because only these functions can possess valid inverses. The relationship between a function and its inverse is direct: if \( f \) is one-to-one, the inverse \( g \) can be uniquely determined, as is the case with \( y = \arcsin x \).
Hence, recognizing and proving a function's one-to-one nature is essential before attempting to find its inverse.

Function Symmetry

Symmetry often simplifies the understanding of function behaviors and their geometric representations. When discussing inverse functions, symmetry plays a significant role, especially across the line \( y = x \), which is also known as the identity line or the \(45^\circ\) line.
For a function and its inverse, their graphs are mirror images across this identity line: if a point \((a, b)\) lies on the function \( f \), then \((b, a)\) will be on the inverse \( g \). This mirroring effect indicates the symmetrical properties of inverse functions across \( y = x \).
In the exercise, \( y = \sin x \) and its inverse \( y = \arcsin x \) showcase this concept. Plotting these functions and their tangent lines uncovers this symmetry, providing insight into their relationship.
Understanding this symmetry is crucial for visualizing how functions transform into their inverses and helps in plotting and analyzing them effectively.