1. ## Proving inverses

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.

2. 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 put the revisions in red. Now when you are trying to show $f(f^{-1}(x))$ you should get:

$y=2-3x$

$y=2-3(\frac{x-2}{-3})$

$y=2+(x-2)$

$y=x$

You would get the exact same answer if you switched which function you were composing the other of.

3. 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.
As long as a is not 0, a/a= 1. "/a" indicates you are dividing not multiplying.