What is the domain and range of $f(x)=sqrt(x^2+4)$ ?

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Answer 1 · 2015-11-04T23:27:32

The domain is easy, as the square makes everything under the root-sign non-negative, so there are no restrictions on $x$ .

In other words domain $-oo < x<+oo$

Since $x^2>=0->x^2+4>=4->sqrt(x^2+4)>=2$

In other words range $2<=f(x)<+oo$

What is the domain and range of f(x)=sqrt(x^2+4)f(x)=sqrt(x^2+4)?