Question

find the inverse function of f...

Answer 1

\[-\sqrt{16-x ^{2}}\] where \[-4\le x \le0\]

Answer 2

i know the answer but i don't know how to do it :(

Answer 3

\[y=-\sqrt{16-x^2}\]
solve for x
\[y^2=16-x^2 =>x^2+y^2=16 =>x^2=16-y^2=>x=\pm \sqrt{16-y^2}\]
but we need to determine if it is positive or negative
think
the domain of f is [-4,0]
and the range of f is ....[-4,0]

so \[f^{-1}(x)=-\sqrt{16-x^2}\]

Answer 4

thank you so much! very clear

Answer 5

i really appreciate the time you took to answer