Question

what the.... uhmmm... kay i'm lost. help please?

Answer 1

If f(x)=2x+1, write the defining equation of \[f(fof^{-1})(x)\]

Answer 2

This is a bother :D
First find \(\large f^{-1}(x)\)

Answer 3

uhhh 1/(2x+1) ??

Answer 4

No... interestingly enough :D
To get the inverse function, what to do is replace f(x) with x and replace the x with y.
Getting
\[\large x = 2y +1\]
and solve for y.

Answer 5

sorry... i haven't had to look for an inverse of a function in a while. haha i forgot.

Answer 6

.... sooooooo if i did this right, it's (x-1)/2 ???

Answer 7

You did :)
So, let's first find
\[\Large (f\circ f^{-1})(x) \]

Answer 8

It's basically
\[\large f[f^{-1}(x)]\]

Answer 9

soo... \[f(\frac{ x-1 }{ 2 })\]

Answer 10

Yes .... so...
\[\Large 2\left(\frac{x-1}{2}\right)+1\]

Answer 11

ahhhhh!!!! omg thanks so much for breaking that down

Answer 12

... is it x?

Answer 13

LOL yeah... I withheld something from you :3

Answer 14

??

Answer 15

\[\Large (f\circ f^{-1})(x) = x\]
composing a function with its inverse ALWAYS results in x... regardless of f :)

Answer 16

But I thought it'd be fun to have you work it out first :)

Answer 17

But yeah, you'd do well to remember that... no matter what function f is...
\[\huge (f\circ f^{-1})(x) = x =(f^{-1}\circ f)(x)\]

Answer 18

that's fine! :) lol at least now i know cause it looked very confusing.

Answer 19

so, in effect...
\[\Large f(fof^{-1})(x)\]
is just \[\Large f(x)\]

Which is just...? :)

Answer 20

x?

Answer 21

No... :/
f(x) is actually given in your question :P

Answer 22

lmaao ... shoot.

Answer 23

2x+1

Answer 24

And there you have it :P

Answer 25

thanks! now i have to learn to decompose a function :S

Answer 26

Remember this
\[\huge (f\circ f^{-1})(x) = x =(f^{-1}\circ f)(x)\]
It might just save you precious minutes of exam time :D

Answer 27

which i have on tuesday!! :O thank you again!