Question

Domain and range of f(x)=square root of (49-x^2)

Answer 1

\(\bf \sqrt{49-x^2}\)
what values for "x" will make the radical quantity 0?

Answer 2

7 and -7. I got confused cause i am not sure whether to look at it as a circle or as a square root function

Answer 3

7 and -7 are correct, so that's your domain, "x" will have to be between -7 and 7

if it becomes bigger than that, you'd end up with a negative inside the radical and that'd make it an imaginary number

Answer 4

whatever is inside of the square root has to be > or = to 0. So
\[49-x ^{2}\ge 0\]
Resolve to get x alone and you'll have the domain.

As for the range, y will always be > or = to what? hint, look at the parent function.

Answer 5

as far the Range, "y", well the radical when "x" becomes 7 or -7 will turn to 0

and when "x" is in between -7 and +7, 49 will be the bigger value and thus positive
how far up positive? well, if we set "x = 0", you'd be left with \(\bf \sqrt{49-0}\)
which yields no more than 7 or -7