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Chapter 2: Problem 18

Find the domain of each function. $$f(x)=\sqrt{x+2}$$

Short Answer

Expert verified
The domain of the function \(f(x)=\sqrt{x+2}\) is \([-2, \infty)\).

Step by step solution

01

Understand the function's domain restriction

For any function involving a square root, the argument of the square root (the expression under the square root symbol) must be greater than or equal to 0. This is because negative numbers do not have real square roots, only imaginary ones. So in this case, \(x + 2\) must be greater than or equal to 0.
02

Solve the inequality

Take the inequality \(x + 2 \geq 0\) and solve it for x. To do this, subtract 2 from both sides of the inequality to isolate x. The solution becomes \(x \geq -2\).
03

Express the domain

Now that we know the inequality that x must satisfy, the domain can be written in interval notation. The domain of the function includes all x-values greater than or equal to -2. Therefore, the domain in interval notation is \([-2, \infty)\).

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Most popular questions from this chapter

How is the standard form of a circle's equation obtained from its general form?

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}-2 x+y^{2}-15=0 $$

graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. $$ \begin{aligned} x^{2}+y^{2} &=9 \\ x-y &=3 \end{aligned} $$

write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(-3,-1), r=\sqrt{3} $$

will help you prepare for the material covered in the next section. $$ \text { If }\left(x_{1}, y_{1}\right)=(-3,1) \text { and }\left(x_{2}, y_{2}\right)=(-2,4), \text { find } \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $$
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