Problem 3
Question
(a) \(f(x)=\sqrt{x}\) (b) \(g(x)=\sqrt{x-3}\) (c) \(h(x)=\sqrt{x^{2}-4}\)
Step-by-Step Solution
Verified Answer
The domain of \(f(x)\) is [0, \( \infty \)), the domain of \(g(x)\) is [3, \( \infty \)), and the domain of \(h(x)\) is (-\( \infty \), -2] U [2, \( \infty \)).
1Step 1: Determine the domain of \(f(x)=\sqrt{x}\)
The square root function, \( f(x) = \sqrt{x} \), is defined for all \( x \geq 0 \). Therefore, the domain of the function \( f(x) \) is [0, \( \infty \)).
2Step 2: Determine the domain of \(g(x)=\sqrt{x-3}\)
For this function, it can only exist when the value inside the square root, \( x - 3 \), is positive or zero. Therefore, \( x - 3 \geq 0 \). Solving for \( x \) gives \( x \geq 3 \). So, the domain for \( g(x) \) is [3, \( \infty \)).
3Step 3: Determine the domain of \(h(x)=\sqrt{x^{2}-4}\)
This function exists when the value inside the square root, \( x^{2} - 4 \), is positive or zero. We can solve \( x^{2} - 4 \geq 0 \) by factoring it to \( (x - 2)(x + 2) \geq 0 \). From the resulting inequality, we can observe that \( h(x) \) is defined for \( x \leq -2 \) or \( x \geq 2 \). Therefore, the domain for \( h(x) \) is (-\( \infty \), -2] U [2, \( \infty \)).
Key Concepts
Square Root FunctionInequalitiesFunction Domain Determination
Square Root Function
The concept of the square root function is central to understanding certain types of algebraic equations. A square root function is expressed as \( f(x) = \sqrt{x} \) and involves finding the number that, when multiplied by itself, gives the original number, \( x \).
For example, since \( 3 \times 3 = 9 \), the square root of \( 9 \) is \( 3 \). This function is inherently linked to the concept of squares and square numbers. But the critical point to note is that a square root function only accepts non-negative inputs, as the square root of a negative number is not defined within the realm of real numbers.
This limitation on input values has a direct effect on the domain of the function, which contains all allowable inputs to the function. For \( f(x) = \sqrt{x} \), the domain is all real numbers greater than or equal to zero, expressed as \([0, \infty)\). Understanding this characteristic is vital when working with any equations involving square roots.
For example, since \( 3 \times 3 = 9 \), the square root of \( 9 \) is \( 3 \). This function is inherently linked to the concept of squares and square numbers. But the critical point to note is that a square root function only accepts non-negative inputs, as the square root of a negative number is not defined within the realm of real numbers.
This limitation on input values has a direct effect on the domain of the function, which contains all allowable inputs to the function. For \( f(x) = \sqrt{x} \), the domain is all real numbers greater than or equal to zero, expressed as \([0, \infty)\). Understanding this characteristic is vital when working with any equations involving square roots.
Inequalities
Inequalities are mathematical expressions that show the relative size or order of two values. They are symbolized by signs like \( > \) (greater than), \(< \) (less than), \(\geq\) (greater than or equal to), and \(\leq\) (less than or equal to).
For instance, if we have an inequality \( x - 3 \geq 0 \), it lets us know that the value of \( x \) must be greater than or equal to \( 3 \) for this inequality to hold true. Solving inequalities is a process similar to solving equations, but the solutions are often ranges of numbers rather than specific values. This is particularly crucial when determining the domain of functions involving square roots, as we need to ensure the values beneath the square root sign are non-negative to identify the domain of validity for the function.
For instance, if we have an inequality \( x - 3 \geq 0 \), it lets us know that the value of \( x \) must be greater than or equal to \( 3 \) for this inequality to hold true. Solving inequalities is a process similar to solving equations, but the solutions are often ranges of numbers rather than specific values. This is particularly crucial when determining the domain of functions involving square roots, as we need to ensure the values beneath the square root sign are non-negative to identify the domain of validity for the function.
Function Domain Determination
Determining the domain of a function is a fundamental step in working with mathematical functions. The domain is the set of all possible inputs for which the function is defined. When dealing with square root functions, such as \( h(x) = \sqrt{x^2 - 4} \), the domain is found by setting the expression inside the square root to be greater than or equal to zero, to avoid taking the square root of a negative number.
For the function \( h(x) \), the expression \( x^2 - 4 \) is factored into \( (x-2)(x+2) \), and solving \( (x-2)(x+2) \geq 0 \) allows us to find the ranges of \( x \) for which this function is defined. Through analyzing the factors, we find that the function has a domain of \( (-\infty, -2] \cup [2, \infty) \). This means that \( h(x) \) is real and valid for all inputs less than or equal to \( -2 \) or greater than or equal to \( 2 \). Grasping domain determination is essential for graphing functions, solving equations, and understanding the behavior of functions across different intervals.
For the function \( h(x) \), the expression \( x^2 - 4 \) is factored into \( (x-2)(x+2) \), and solving \( (x-2)(x+2) \geq 0 \) allows us to find the ranges of \( x \) for which this function is defined. Through analyzing the factors, we find that the function has a domain of \( (-\infty, -2] \cup [2, \infty) \). This means that \( h(x) \) is real and valid for all inputs less than or equal to \( -2 \) or greater than or equal to \( 2 \). Grasping domain determination is essential for graphing functions, solving equations, and understanding the behavior of functions across different intervals.
Other exercises in this chapter
Problem 2
(a) \(f(x)=\frac{1}{x+2}\) (b) \(g(x)=\frac{5}{x^{2}+4}\)
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Which of the following rules can be modeled as a function? If a rule is not a function, explain why not. (a) For a particular flask, the rule assigns to every v
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(a) \(f(x)=\sqrt{\frac{1}{x+1}}\) (b) \(g(x)=\sqrt{\frac{x}{x+1}}\)
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There are infinitely many prime numbers. This has been known for a long time; Euclid proved it sometime between 300 B.C. and 200 B.C. \({ }^{6}\) Number theoris
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