Question

Can someone help me with this

Answer 1

want to find the domain and range first, or the inverse first?

Answer 2

inverse please @satellite73

Answer 3

ok lets start with
\[y=x^2-2\] switch \(x\) and \(y\) to get
\[x=y^2-2\] and solve for \(y\) in two steps

Answer 4

add 2, get \[x+2=y^2\] take the square root (don't forget the \(\pm\) get \[y=\pm\sqrt{x+2}\] so the inverse is not a function, since it has the \(\pm\) in front

Answer 5

you ok with that?

Answer 6

okay yes yes i got that

Answer 7

ok now for the domain and range
lets find the domain and range of \(f(x)=x^2-2\) first

Answer 8

since \(f\) is a polynomial, the domain is all real numbers \(\mathbb{R}\)
since a square is never negative, the range is \([-2,\infty)\) or \(y\geq -2\) however you like to write it

Answer 9

to find the domain and range of the inverse, switch those two
so the domain of the inverse is \([-2\infty)\) and the range is \(\mathbb{R}\)

Answer 10

oops i meant \[[-2,\infty)\] but you get the idea

Answer 11

okay thank you so much

Answer 12

yw