You said that (xy) = 16, and (x+y) = 4 .
From the second equation you can get [ x = 4 - y ],
then substitude that for 'x' in the first equation, and
finally, rearrange the first equation to read
<u>x² - 4x + 16 = 0</u>
Don't even try to factor that quadratic equation. Go straight
to the quadratic formula, and the two solutions you find are ...
<em>x = 2 + i 2√3</em>
and
<em>x = 2 - i 2√3</em> .
Those are the two number that do what you want.
There are no <u>real</u> numbers that can do it.
we know that f(1)=2 so f^-1(2)=1
so B is correct!
