Problem 36 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=\left(\sin ^{-1} x\right)\left(\cos ^{-1} x\right) \text { on }[0,1]$$

Step-by-Step Solution

Answer

Question: Determine the critical point of the function \(f(x) = (\sin^{-1}(x))(\cos^{-1}(x))\) on the interval [0, 1] and identify whether it corresponds to a local maximum, local minimum, or neither. Answer: The critical point of the function \(f(x) = (\sin^{-1}(x))(\cos^{-1}(x))\) on the interval [0, 1] is \(x = \frac{\sqrt{2}}{2}\), which corresponds to a local maximum.

Step 1: Find the derivative of the function

To find the derivative, we need to use the product rule which states that if you have a function \(h(x) = g(x) * m(x)\), then \(h'(x) = g'(x) * m(x) + g(x) * m'(x)\). In our case, let \(g(x) = \sin^{-1}(x)\) and \(m(x) = \cos^{-1}(x)\). Now, let's find the first derivative of each: \(g'(x) = \frac{1}{\sqrt{1 - x^2}}\) \(m'(x) = -\frac{1}{\sqrt{1 - x^2}}\) Now, we will apply the product rule: \(f'(x) = g'(x) * m(x) + g(x) * m'(x) = \frac{1}{\sqrt{1 - x^2}}(\cos^{-1}(x)) + (\sin^{-1}(x))\left(-\frac{1}{\sqrt{1 - x^2}}\right)\)

Step 2: Set the derivative equal to zero and solve for x

Now that we have the derivative, we set it equal to zero to find the critical points: \(f'(x) = 0 \implies \frac{1}{\sqrt{1 - x^2}}(\cos^{-1}(x)) + (\sin^{-1}(x))\left(-\frac{1}{\sqrt{1 - x^2}}\right) = 0\) Rearrange the equation and notice that both terms have a common factor of \(\frac{1}{\sqrt{1 - x^2}}\), so we can divide the entire equation by that factor, which will simplify the equation: \((\cos^{-1}(x)) - (\sin^{-1}(x)) = 0\) Now, we need to solve for x: \(\cos^{-1}(x) = \sin^{-1}(x)\) Using the identity \(\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)\), we get: \(\cos^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)\) \(2\cos^{-1}(x) = \frac{\pi}{2}\) Now, we can find x: \(x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)

Step 3: Use a graphing utility to determine the type of critical point

Using a graphing utility, plot the function \(f(x) = (\sin^{-1}(x))(\cos^{-1}(x))\) on the interval [0, 1]. Observe the behavior of the function around the critical point \(x = \frac{\sqrt{2}}{2}\). Upon examining the graph, we can see that the function reaches a clear local maximum at the critical point \(x = \frac{\sqrt{2}}{2}\).

Step 4: Conclusion

The critical point of the function \(f(x) = (\sin^{-1}(x))(\cos^{-1}(x))\) on the interval [0, 1] is \(x = \frac{\sqrt{2}}{2}\), which corresponds to a local maximum.

Key Concepts

Product Rule

The Product Rule is a fundamental technique in calculus used to find the derivative of a product of two functions. This is particularly important when dealing with functions that are composed together, as it allows us to differentiate them accurately. The formula for the Product Rule is:

\[ h'(x) = g'(x) \, m(x) + g(x) \, m'(x) \]

Here, if \( h(x) = g(x) \cdot m(x) \), then \( g(x) \) and \( m(x) \) are the original functions, and \( g'(x) \) and \( m'(x) \) are their respective derivatives. In the exercise, we encountered \( g(x) = \sin^{-1}(x) \) and \( m(x) = \cos^{-1}(x) \). By applying the product rule, we calculate the derivative of these functions' product. Understanding the product rule is essential as it simplifies the process of finding critical points and analyzing the behavior of complex expressions.

Inverse Trigonometric Functions

Inverse Trigonometric Functions, such as \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \), play a critical role in calculus, especially when solving trigonometric equations. These functions provide the angle whose trigonometric function equals a given number. For example, \( \sin^{-1}(x) \) returns the angle whose sine is \( x \). These functions have specific domains and ranges:

• \( \sin^{-1}(x) \) is defined for \( -1 \leq x \leq 1 \) and returns angles in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
• \( \cos^{-1}(x) \) is also defined for \( -1 \leq x \leq 1 \) but returns angles in \( [0, \pi] \).

When solving \( \cos^{-1}(x) = \sin^{-1}(x) \) in the exercise, we utilized an identity that connects these functions to simplify the solution. Mastery of these concepts is crucial for successfully managing integrals and derivatives that involve these inverse operations.

Graphing Utility

A Graphing Utility is a tool that helps visualize functions and their derivatives, making it easier to identify key features of graphs such as critical points, intercepts, and asymptotes. It provides a visual representation which can be invaluable in understanding the behavior of functions. In the context of the exercise, a graphing utility can help verify where the function \( f(x) = (\sin^{-1}(x))(\cos^{-1}(x)) \) reaches its critical point by letting us see the function's graph. By observing the graph, especially around the critical point \( x = \frac{\sqrt{2}}{2} \), it becomes clear whether this point corresponds to a local maximum, minimum, or neither. Utilizing graphing utilities strengthens comprehension by providing a visual aid that supplements analytical work.

Local Maximum

A Local Maximum is a point on a graph where a function reaches its highest value within a particular region, regardless of what's happening further out on the graph. It indicates a peak in the function's behavior, and it's found by analyzing the first and second derivatives. In the exercise, the critical point \( x = \frac{\sqrt{2}}{2} \) on the function \( f(x) = (\sin^{-1}(x))(\cos^{-1}(x)) \) was identified as a local maximum using a graphing utility. This determination was made by analyzing the graph's trend and confirming that the value of \( f(x) \) decreases after this critical point within the given interval. Recognizing local maxima is important as it provides insight into the optimization and behavior of various mathematical and applied problems.