f(x)=5x^2-1 find the inverse and any restrictions on the domain and range, showing all steps. plz help!!!

Question

anonymous · Answer

nick could u help plz?

anonymous · Answer

y=5x^2-1
sqrt{(y+1)/1}=x

y.>-1

anonymous · Answer

@ash2326 do you know how to do this? I don't remember it...at all.

ash2326 · Answer

@AbbeyD16 
we have
$$f(x)=5x^2-1$$
we need to find inverse of f(x) that's $f^{-1}(x)$
one important thing to keep in mind
$$f(f^{-1}(x))=f^{-1} (f (x))=x$$
Do you get this point ?

anonymous · Answer

yea

anonymous · Answer

Thank you so much, Ash!

ash2326 · Answer

@rebeccaskell94 we've just started
now let's put $x=f^{-1} (x)$ in f(x)
so
$$f(x)=5x^2-1$$
$$f(f^{-1} x)=5(f^{-1} x)^2-1$$
@AbbeyD16 what'd be the left side?

anonymous · Answer

Good luck, Abbey! I know Ash will do a great job with helping you! :D

anonymous · Answer

im not sure :/

ash2326 · Answer

@AbbeyD16 Sorry I was disconnected, the answer is in my posts. Please read them again

anonymous · Answer

u there?

ash2326 · Answer

yes

anonymous · Answer

1. Solve x in terms of f(x)
$$y=5x^2-1$$
$$y+1=5x^2$$
$$x^2=(y+1)/5$$
$$x=\sqrt{(y+1)/5}$$
Thus the inverse function $$ƒ^{−1}$$ is given by the formula
$$f^{-1}(y)= \sqrt{(y+1)/5}$$