Originally Posted by

**fezz349** Work in groups of two or three to

a. find f^-1 and show that (f of f^-1)(x)= (f^-1 of f)(x) = x

b. graph f and f^-1 in the same viewing window.

f(x)=2-3x

note: the of is in replacement of the that little unfilled circe that's supposed to be between f and f^-1, and f^-1 and f.

Here's what i did so far and where I got stuck:

y=2-3x

x=2-3y

x-2=3y

y=(x-2)\3

I now found my inverse equation by switching the x and y values and solving.

Now when I'm trying to prove that (f of f^-1)(x)= (f^-1 of f)(x) = x ... you have to put the equations into each other and come out with an answer of x, so...

y=2-3x

y=(x-2)/3

2-3(x-2/3)

2-3x+6 ... now what happens when I multiply the -3 with the 3 in the denominator, does it come out to be -9 or do they cancel out to a negative? I'm lost from that point on.