How do you find domain and range for $y = \sqrt{x^{2} + 4}$ $y=sqrt(x^2+4)$ ?

Question

How do you find domain and range for $y = \sqrt{x^{2} + 4}$ $y=sqrt(x^2+4)$ ?

1 Answer

Impact of this question

12357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-12T02:50:37

Domain: $RR$
Range: $[2,+oo)$

Explanation:

Domain
The first thing you should consider is that the number inside the radical should be nonnegative so it won't be imaginary. We can compute for the domain by solving the inequality $x^2+4>=0$ .

$color(white)(XX)x^2+4>=0$
$color(white)(XX)x^2>=-4$

Since any real number you substitute into $x$ is squared, it will always be nonnegative and greater than $-4$ . Therefore the domain is $RR$ .

Range
The lowest value the radical can have is $sqrt4$ or $2$ (when $x=0$ ). Therefore, the range is $y>=2$ or $[2,+oo)$ .

Science

Math

Humanities

... and beyond

How do you find domain and range for $y = \sqrt{x^{2} + 4}$ $y=sqrt(x^2+4)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find domain and range for y=sqrt(x^2+4)y=√x2+4y=sqrt(x^2+4)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find domain and range for $y = \sqrt{x^{2} + 4}$ $y=sqrt(x^2+4)$ ?