Finding the range, radical in the denominator

Mar 2014
61
0
Canada
Hi,

I am trying to find the range of

f(x) = 1 / [sqrt(x+3)]

I know that the function is undefined when x+3 =< 0 so the domain is all x>-3
As the denominator must be positive, i know that f(x) > 0 and i believe that is the range.

I am having trouble knowing how I should set this out on paper / what my working should look like.

Any help would be appreciated, thanks.
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
Hi,

I am trying to find the range of

f(x) = 1 / [sqrt(x+3)]

I know that the function is undefined when x+3 =< 0 so the domain is all x>-3
As the denominator must be positive, i know that f(x) > 0 and i believe that is the range.
At one end of it's domain $f(x) \to \infty$ as $x \to -3$ from above.

At the other end of it's domain $f(x) \to 0$ as $x \to \infty$

So the range of $f(x)$ is $(0,\infty)$
 
Dec 2012
1,145
502
Athens, OH, USA
Hi,
As you know, the range of a function f with domain D is the set $\{f(x):x\in D\}$. So for $f(x)={1\over\sqrt{x+3}}$ with $D=(-3,\infty)$, you think the range is $(0,\infty)$.

You already observed that for $x\in D$, $f(x)\in(0,\infty)$. You're halfway there. Let $b\in(0,\infty)\text{ or }b>0$. Then set $x=-3+{1\over b^2}\in D$. Then $f(x)={1\over\sqrt{3+-3+1/b^2}}=\sqrt{b^2}=b$ and so $b$ is in the range.

How did I magically choose x? Why I solved the equation $f(x)=b$ for x. Note however, that solving this equation for x does not say b is in the range until I verify this as above.