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Chapter 1: Problem 6

Find the natural domain of the function algebraically, and confirm that your result is consistent with the graph produced by your graphing utility. [Note: Set your graphing utility to the radian mode when graphing trigonometric functions.] (a) \(f(x)=\sqrt{3 x-2}\) (b) \(g(x)=\sqrt{9-4 x^{2}}\) (c) \(h(x)=\frac{1}{3+\sqrt{x}}\) (d) \(G(x)=\frac{3}{x}\) (e) \(H(x)=\sin ^{2} \sqrt{x}\)

Short Answer

Expert verified
(a) \([\frac{2}{3}, \infty)\); (b) \([-\frac{3}{2}, \frac{3}{2}]\); (c) \([0, \infty)\); (d) \((-\infty, 0) \cup (0, \infty)\); (e) \([0, \infty)\)."

Step by step solution

01

Determine Domain of Function (a)

For the function \( f(x) = \sqrt{3x-2} \), the expression inside the square root, \( 3x-2 \), must be non-negative. Set up the inequality: \( 3x - 2 \geq 0 \). Solve this inequality: \( 3x \geq 2 \), therefore \( x \geq \frac{2}{3} \). Thus, the domain of \( f(x) \) is \( [\frac{2}{3}, \infty) \).
02

Determine Domain of Function (b)

For the function \( g(x) = \sqrt{9 - 4x^2} \), the expression under the square root, \( 9 - 4x^2 \), must be non-negative. Set the inequality \( 9 - 4x^2 \geq 0 \) which becomes \( 9 \geq 4x^2 \). Solving, we have \( \frac{9}{4} \geq x^2 \), implying \( -\frac{3}{2} \leq x \leq \frac{3}{2} \). Hence, the domain of \( g(x) \) is \( [-\frac{3}{2}, \frac{3}{2}] \).
03

Determine Domain of Function (c)

In the function \( h(x) = \frac{1}{3+\sqrt{x}} \), the denominator \( 3 + \sqrt{x} \) must not be zero. Hence, \( 3 + \sqrt{x} eq 0 \), which implies \( \sqrt{x} eq -3 \). Since \( \sqrt{x} \) is always non-negative, we do not need any further restrictions beyond \( x \geq 0 \). Therefore, the domain of \( h(x) \) is \( [0, \infty) \).
04

Determine Domain of Function (d)

For \( G(x) = \frac{3}{x} \), the expression in the denominator, \( x \), must not be zero. Thus \( x eq 0 \), meaning the domain of \( G(x) \) is \( (-\infty, 0) \cup (0, \infty) \).
05

Determine Domain of Function (e)

The function \( H(x) = \sin^2(\sqrt{x}) \) involves a square root and a sine function. For \( \sqrt{x} \) to be real, \( x \) must be non-negative, hence \( x \geq 0 \). The sine squared function is defined for all real inputs, so no additional restrictions apply. Therefore, the domain of \( H(x) \) is \( [0, \infty) \).

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Key Concepts

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

Trigonometric Functions
Trigonometric functions play a crucial role in various mathematical contexts, particularly in domains involving angles and periodic phenomena. In functions, these include sine, cosine, and tangent, among others. When working with trigonometric functions, the input values (usually angles) affect the function's behavior. For natural domains, it's important to consider angles measured in radians for consistency, especially when using graphing utilities to visualize these functions. Unlike some algebraic functions, trigonometric functions often have comprehensive natural domains but may sometimes require combining with other functions, affecting overall domain requirements.
Inequalities
Inequalities are powerful tools in determining the permissible input range, or domain, of a function. By setting up expressions such as \(3x - 2 \ge 0\), as seen in the step-by-step solution, one can determine the values of \(x\) that make the function valid. Solving inequalities frequently involves rearranging terms and dividing by coefficients to isolate the variable of interest. The results of inequalities directly inform the domain of a function, indicating where a function is defined and where it might encounter division by zero, negative values under a square root, or other difficulties.
Graphing Utilities
Graphing utilities, such as graphing calculators or software, offer a visual confirmation of the algebraic domain findings. They allow for plotting functions to observe their behavior over specified ranges. When graphing functions like these, it's vital to ensure the utility is set to 'radian mode' for trigonometric functions to ensure accuracy. By examining function plots, one can corroborate the predetermined domain, noticing where the graph stops or exhibits discontinuities. This visual approach complements the algebraic method by providing tangible evidence of function behavior and domain limitations.
Domain in Calculus
In calculus, understanding the domain of a function is fundamental. The domain indicates the set of all input values for which a function is defined. Without a clear grasp of this, operations like differentiation or integration could be misleading or impossible. For functions like \(h(x) = \frac{1}{3+\sqrt{x}}\), the domain arises from ensuring non-zero denominators and real values under roots. It's vital to assess these conditions as they ensure the function's applicability in calculus-based applications. Evaluating domains requires checking constraints brought by each mathematical element in a function, paving the way for further exploration using calculus.

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