Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the average rate of change of \(f(x)=\sin ^{-1} x\) from \(x=-1\) to \(x=1\)

Short answer

The average rate of change is \( \frac{\text{π}}{2} \).

Step-by-step solution

- Understand the Formula for Average Rate of Change

The average rate of change of a function between two points is given by \[\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}\]. Here, the function is \( f(x) = \text{arcsin}(x) \) and the interval is from \( x = -1 \) to \( x = 1 \).

- Evaluate the Function at Endpoints

Calculate \( f(-1) \) and \( f(1) \): \[ f(-1) = \text{arcsin}(-1) = -\frac{\text{π}}{2} \]\[ f(1) = \text{arcsin}(1) = \frac{\text{π}}{2} \]

- Apply the Average Rate of Change Formula

Substitute \( f(-1) \) and \( f(1) \) into the formula: \[ \text{Average Rate of Change} = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{\frac{\text{π}}{2} - \big(-\frac{\text{π}}{2}\big)}{1 - (-1)} = \frac{\frac{\text{π}}{2} + \frac{\text{π}}{2}}{2} = \frac{\text{π}}{2} \]

Key concepts

arcsin function

The arcsin function, also known as the inverse sine function, is written as \( \sin^{-1}(x) \) or \( \text{arcsin}(x) \). This function is the inverse of the sine function and returns the angle whose sine is \( x \).

It is important to know that the range of the arcsin function is limited to \(-\frac{\pi}{2} to \frac{\pi}{2}\). This means the output values (the angle \( y \)) will always fall within this interval.

For example, if \( \sin(y) = x \), then \( y = \sin^{-1}(x) \).

In our original problem, we used the arcsin function to evaluate the function \( f(x) = \sin^{-1}(x) \), where \( x = -1 \) and \( x = 1 \).

evaluating trigonometric functions

Evaluating trigonometric functions involves finding their values at specific points. For our problem, we need to calculate \( f(-1) \) and \( f(1) \) for \( f(x) = \text{arcsin}(x) \).

For \( x = -1 \):

• The sine of an angle that results in -1 is \( -\frac{\pi}{2} \).
• Therefore, \( f(-1) = \text{arcsin}(-1) = -\frac{\pi}{2} \).

Now for \( x = 1 \):

• The sine of an angle that results in 1 is \( \frac{\pi}{2} \).
• Therefore, \( f(1) = \text{arcsin}(1) = \frac{\pi}{2} \).

Understanding these basic evaluations will help in solving more complex problems related to trigonometric functions.

interval calculations in calculus

One important concept in calculus is finding the average rate of change of a function over an interval. This involves evaluating the function at both endpoints of the interval and applying the average rate of change formula: \(\frac{f(b) - f(a)}{b - a}\).

Let's consider our problem again where we need to find the average rate of change of \( f(x) = \text{arcsin}(x) \) from \( x = -1 \) to \( x = 1 \). We have already calculated:

• \( f(-1) = -\frac{\pi}{2} \)
• \( f(1) = \frac{\pi}{2} \)

Now we can apply the formula:

• First, we find the difference in function values: \( f(1) - f(-1) = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi \).
• Next, we find the difference in \( x \) values: \( 1 - (-1) = 2 \).
• Finally, we find the average rate of change: \(\frac{\pi}{2}\).

This tells us that the average rate of change of \( f(x) = \text{arcsin}(x) \) from \( x = -1 \) to \( x = 1 \) is \( \frac{\pi}{2} \).