Answer:
(f^−1)′(6)=1/(f'(f^-1(6)))
(f^−1)′(6)=1/(f'(f^-6)))
I hope this helps.
Let x and y be the two numbers. We have:

Subtract the first equation from the second to get

And deduce

The two numbers are 3 and 11.
The distance between is (8,6).
(3x+2) (x-7)
FOIL
first 3x*x = 3x^2
outer 3x*-7 = -21x
inner = 2*x = 2x
last 2*-7 = -14
add them together = 3x^2 -21x+2x -14
combine like terms 3x^2 -19x -14