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Chapter 1: Problem 20

Find the domain of the function. $$ h(x)=\frac{1}{\sin x-\frac{1}{2}} $$

Short Answer

Expert verified
The domain of the function \( h(x) = \frac{1}{\sin x-\frac{1}{2}} \) is all real numbers except \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.

Step by step solution

01

Identify the Critical Point

To solve this problem, first set the denominator of the function equal to zero and solve the equation for x. Thus, the equation will be \(\sin x-\frac{1}{2} = 0\)
02

Solve for x

Re-arrange and solve for x, the equation now becomes \(\sin x = \frac{1}{2}\). Solve this equation for all possible solutions. The solutions are \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.
03

Define the Function Domain

The solution in step 2 represents the values at which the denominator is zero, and thus, these values need to be excluded from the domain of the function \(h(x)\). Therefore, the function domain will be all real numbers except \(x = \frac{\pi}{6} + 2\pi k\) and \(x = \frac{5\pi}{6} + 2\pi k\), where k is an integer.

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are a fundamental part of mathematics, often dealing with angles and periodic phenomena. They relate angles of a triangle to the lengths of its sides via functions like sine, cosine, and tangent. In this problem, we focus on the sine function.
  • The sine function, denoted as \( \sin x \), takes an angle \( x \) (usually in radians), and outputs a value between -1 and 1.
  • This function is periodic with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) radians.
  • Sine is positive in the first and second quadrants (0 to \( \pi \)), negative in the third and fourth quadrants (\( \pi \) to \( 2\pi \)).
Understanding these aspects of sine is crucial for finding where \( \sin x = \frac{1}{2} \), as the sine of certain angles corresponds to these fractional values.
Undefined Values
Functions can have values that are undefined, something we often need to watch out for. In our function \( h(x)=\frac{1}{\sin x-\frac{1}{2}} \), undefined values occur where the denominator equals zero.

For the given equation, the denominator is \( \sin x - \frac{1}{2} \). To find when this expression is zero, we solve \( \sin x = \frac{1}{2} \).
  • The angles satisfying this equation in standard positions are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
  • Because sine is periodic, these solutions repeat every \( 2\pi \), resulting in an infinite number of solutions: \( x = \frac{\pi}{6} + 2\pi k \) and \( x = \frac{5\pi}{6} + 2\pi k \), where \( k \) is any integer.
When calculating domain, these x-values must be excluded, otherwise the function becomes undefined.
Continuous Function
Continuous functions allow us to map inputs to outputs without interruption. The function \( h(x) \) itself needs a continuous domain wherever it doesn鈥檛 face division by zero.
A function is continuous if there are no breaks, jumps, or holes in its graph for the set of inputs considered.

Since the function \( h(x)=\frac{1}{\sin x-\frac{1}{2}} \) is made up from sine, a continuous function, we consider continuity minus the excluded points identified earlier.
  • The function is continuous everywhere sine is defined, except at \( x = \frac{\pi}{6} + 2\pi k \) and \( x = \frac{5\pi}{6} + 2\pi k \), excludes where it causes division by zero.
  • Between these points, \( h(x) \) behaves predictably without disruption, showing smooth transitions across its domain.
This ensures that \( h(x) \) is continuous for its valid input values, providing a complete understanding of its graph behavior.

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Most popular questions from this chapter

Prove that for any real number \(y\) there exists \(x\) in \((-\pi / 2, \pi / 2)\) such that \(\tan x=y\)

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{2}-6 x+8, \quad[0,3], \quad f(c)=0 $$

Use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval [0, 1]. Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ g(t)=2 \cos t-3 t $$

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In Exercises \(25-34,\) find the limit. $$ \lim _{x \rightarrow 1 / 2} x \sec \pi x $$
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