How to find range inside square root

[Mohmmad Maaitah]

Homework Statement

find range

Relevant Equations

none

Hi, so I know how to find domain but how about range in this problem?
I don't understand the way he did it?
I always get answers wrong when it comes to range.

 

[anuttarasammyak]

Homework Statement: find range
Relevant Equations: none

Hi, so I know how to find domain but how about range in this problem?
I don't understand the way he did it?

Say domain x $\neq$ 2 is all right.
f(x)=\sqrt{x+2}\frac{\sqrt{x-2}}{\sqrt{x-2}}
For x $\neq$ 2 it is simply
f(x)=\sqrt{x+2}
This is monotonically increasing function from 0 to infinity for -2<x<+$\infty$ when we forget x $\neq$ 2.
So $\sqrt{2+2}=2$ is the only one positive value which is out of range.

 

[FactChecker]

In determining the range, the shape of the function is important. The square root function is always increasing, smaller values have smaller square roots and larger values have larger square roots. Therefore, the range is determined by the values at the smallest $x+2$ value ($x+2=0$) and the largest $x+2$ value ($x+2 \rightarrow \infty$). If it wasn't like that, you would have to do more work to determine the maximum and minimum of the function.

You must also keep track of the fact that $x \ne 2$, so the range does not include WRONG:$\sqrt{2+2} = 2\sqrt{2}$. CORRECTION: The range does not include $\sqrt{2+2} = 2$.

 

[Mark44]

You must also keep track of the fact that $x \ne 2$, so the range does not include $\sqrt{2+2} = 2\sqrt{2}$.

You have a typo here.

 

[FactChecker]

You have a typo here.

Thanks! Not a typo, just a brain cell died. I corrected it. :-)