/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Finding Values (a) Use a graph... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Finding Values (a) Use a graphing utility to evaluate arcsin(arcsin 0.5\()\) and arcsin (arcsin \(1 ) .\) (b) Let \(f(x)=\arcsin (\arcsin x)\) Find the values of \(x\) in the interval \(-1 \leq x \leq 1\) such that \(f(x)\) is a real number.

Short Answer

Expert verified
The values for the function f(x) exists for x in the interval -1<=x<=1 and for part (a), arcsin(arcsin(0.5)) = 0.5 and arcsin(arcsin(1)) is undefined.

Step by step solution

01

Evaluate Using a Graphing Utility

Using a graphing utility, one can calculate the arcsine of given numbers. Note, for arcsin(0.5), the arcsine is the angle value whose sine is 0.5. Likewise for arcsin(1), the arcsine is the angle whose sine is 1. Remember to set the mode of your calculator to radians. You will find arcsin(arcsin(0.5)) = 0.5, and arcsin(arcsin(1)) is undefined because the range of arcsin function is [-1,1] and 1 is out of its domain. So the calculator will return an error for this case.
02

Find Real Values for the Function

To find the real values for x in the interval -1<=x<=1 for which f(x) is a real number, we need to consider the domain and range of the arcsine function. Since f(x) = arcsin(arcsin(x)), we apply the arcsin function twice. The domain of the arcsine function is -1<=x<=1 and it ranges between -π/2 and π/2. The inner arcsin(x) gives a number -π/2<=y<= π/2. So, the outer arcsin(y) is real only if -1<=y<=1. But y=arcsin(x) is between -π/2 and π/2, so -1<=y<=1 is always true. So for any x in the interval -1<=x<=1 will make f(x) a real number.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are used to find angles when the result of the trigonometric function is known. These functions essentially reverse the process of the standard trigonometric functions. For example, the arcsine function, denoted as \( \arcsin(x) \), finds the angle \( \theta \) such that \( \sin(\theta) = x \).
This operation is essential in applications where you start from a trigonometric ratio, such as sine, and want to find the corresponding angle. It is important to remember that the principal value of \( \arcsin(x) \) falls within the interval \([-\pi/2, \pi/2]\). The restricted output ensures that each input results in a unique output, making the inverse function a well-defined and properly functional inverse relationship.
In practical terms, \( \arcsin(x) \) helps find the specific angle in a right triangle given the opposite side over the hypotenuse ratio equals \( x \). This function is prevalent in fields such as engineering, physics, and computer graphics.
Domain and Range
Understanding the domain and range of a function is crucial in determining where a function is defined and what values it can achieve. The domain refers to all possible input values, while the range consists of all possible output values. For the arcsine function \( \arcsin(x) \), the domain is \([-1, 1]\) because the sine of an angle cannot exceed these bounds.
In contrast, its range is \([-\pi/2, \pi/2]\), meaning the output, or angle, will always fall within these limits. When dealing with nested functions like \( \arcsin(\arcsin(x)) \), knowing the domain and range of both the inner and outer functions is crucial.
In this scenario, since \( \arcsin(x) \) itself yields results between \([-\pi/2, \pi/2]\), the second application needs to operate within this angle range, verifying that it matches its expected domain quickly, yielding a real number result whenever evaluated correctly.
Graphing Utility
Graphing utilities are powerful tools that can evaluate complex expressions, plot functions, and explore mathematical concepts visually. When you need to evaluate inverse trigonometric functions, like \( \arcsin(x) \), they can compute values quickly and help verify the behavior of these functions.
Using a graphing utility, you can visually understand the effects of applying multiple trigonometric operations. For instance, when asked to calculate \( \arcsin(\arcsin(0.5)) \), the tool shows that this operation produces 0.5 again, as the graph reaffirms \( 0.5 \) is within the valid domain and range circle.
However, attempts to compute expressions outside the defined bounds, such as \( \arcsin(\arcsin(1)) \), result in errors. This is because the arcsine's defined range conflicts with the subsequent arcsine application, which expects input within \([-1, 1]\) but receives an output outside this spectrum from the first call, handing an informative example of range-based limitations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using Properties of Exponents In Exercises \(107-110\) , find the exact value of the expression. \(5^{1 / \ln 5}\)

Solving an Equation Find, to three decimal places, the value of \(x\) such that \(e^{-x}=x\) . (Use Newton's Method or the zero or root feature of a graphing utility.)

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result of part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$\frac{d v}{d t}=-\left(32+k v^{2}\right)$$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part \((c)\) for \(k=0.001 .\) Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral $$\int_{0}^{t_{0}} v(t) d t$$ where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Guidelines for lntegration Describe two ways to alter an integrand so that it fits an integration formula.

Finding an Indefinite Integral In Exercises \(69-76,\) find the indefinite integral. $$\int\left(x^{2}+2^{-x}\right) d x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.