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Chapter 2: Problem 14

Given \(g(x)=\sqrt{x-5},\) the domain is restricted so that \(x \geq\) _____.

Short Answer

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Step by step solution

01

- Understand the Function

Consider the function given as: \[ g(x) = \sqrt{x - 5} \]This is a square root function.
02

- Identify the Domain of the Square Root Function

For the square root function to be defined, the expression inside the square root must be non-negative. Therefore, set up the inequality:\[ x - 5 \geq 0 \]
03

- Solve the Inequality

Solve for \(x\) by adding 5 to both sides of the inequality:\[ x - 5 \geq 0 \]Adding 5:\[ x \geq 5 \]
04

- State the Domain

Based on the solved inequality, the domain of the function \( g(x) = \sqrt{x - 5} \) is all \(x\) such that \( x \geq 5 \).

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Key Concepts

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

square root function
A square root function involves taking the square root of a variable or an expression. It is generally expressed as \(\sqrt{f(x)}\). For example, in our given function \(g(x) = \sqrt{x - 5}\), the variable or expression inside the square root is \(x - 5\).
  • The square root function is only defined for values where the expression inside the square root is non-negative (i.e., zero or positive).
This means the expression inside the square root must always be greater than or equal to zero to give a real number output.
If the expression inside the square root is negative, the output is not a real number.
inequalities
Inequalities are mathematical statements that compare two values or expressions. In our context, inequalities are used to find the domain of the function.
For instance, we started with the inequality \(x - 5 \geq 0\) to ensure the square root function is defined.
Here鈥檚 a step-by-step approach to working with inequalities:
  • First, set the expression inside the square root greater than or equal to zero.
  • Then, solve the inequality to find the range of permissible values for the variable.
By solving \(x - 5 \geq 0\), we added 5 to both sides to obtain \(x \geq 5\). This tells us that any value of \(x\) greater than or equal to 5 will make the function \(g(x) = \sqrt{x - 5}\) valid and produce a real number.
domain restriction
Domain restriction refers to limiting the set of input values for which a function is defined. For our square root function \(g(x) = \sqrt{x - 5}\), the domain is restricted because the value inside the square root must be non-negative.
By solving the inequality \(x - 5 \geq 0\), we concluded that
\( x \geq 5\). Thus, the domain of \(g(x)\) is restricted to all values of \(x\) that are greater than or equal to 5.
Here are some key points to remember when finding the domain restriction for square root functions:
  • Set the expression inside the square root greater than or equal to zero.
  • Solve the resulting inequality to find the permissible values for \(x\).
  • State the domain based on these values.
By understanding domain restrictions, you can better grasp where certain functions are defined and valid.

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Most popular questions from this chapter

Refer to the functions \(r, p,\) and \(q .\) Evaluate the function and write the domain in interval notation. \(r(x)=-3 x \quad p(x)=x^{2}+3 x \quad q(x)=\sqrt{1-x}\) $$(r \cdot q)(x)$$

Determine if the function is even, odd, or neither. $$ r(x)=\sqrt{81-(x+2)^{2}} $$

Refer to the functions \(f\) and \(g\) and evaluate the functions for the given values of \(x\). \(f=\\{(2,4),(6,-1),(4,-2),(0,3),(-1,6)\\} \quad\) and \(\quad g=\\{(4,3),(0,6),(5,7),(6,0)\\}\) $$(f \circ g)(5)$$

Determine if the function is even, odd, or neither. $$ q(x)=\sqrt{16+x^{2}} $$

Given \(f(x)=\frac{12}{x}\) a. Find the difference quotient (do not simplify). b. Evaluate the difference quotient for \(x=2,\) and the following values of \(h: h=0.1, h=0.01, h=0.001\), and \(h=0.0001\). Round to 4 decimal places. c. What value does the difference quotient seem to be approaching as \(h\) gets close to \(0 ?\)
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