# Inverse funciton

• October 27th 2009, 09:12 AM
Oasis1993
Inverse funciton
The function f(x)= ax+b , f-1(x)=8x-3. Find a and b

I would be very happy if you explained how to do it.

Thank you
• October 27th 2009, 09:25 AM
ramiee2010
Quote:

Originally Posted by Oasis1993
The function f(x)= ax+b , f-1(x)=8x-3. Find a and b

I would be very happy if you explained how to do it.

Thank you

let f(x)= ax+b =y (say)
then ax+b =y $\quad \Rightarrow \quad x= \frac{(y-b)}{a}$
$or\ x=\frac{1}{a}y-\frac{b}{a}$
$\therefore f^{-1} (y)=\frac{1}{a}y-\frac{b}{a} \quad \Rightarrow \quad (\because f(x)=y , taking inverse both side )$
$or\ \color {blue} f^{-1} (x)=\frac{1}{a}x-\frac{b}{a}$................(1)
but given $f^{-1}(x)=8x-3$................(2)
on comparison eq(1) and (2)
$\frac{1}{a}=8 \quad \Rightarrow a= \frac{1}{8}$
and
$\frac{b}{a}=3 \quad \Rightarrow b=3a={3\over 8}$
• October 27th 2009, 11:30 AM
Oasis1993
Thank you very much!
• October 28th 2009, 07:52 AM
HallsofIvy
Just to put in my oar: 8x- 3 says "first multiply x by 8 and then subtract 3".

The "inverse" is the inverse of the original operarations ("divide by 8" rather than "multiply by 8" and "add 3" rather than "subtract 3") done in the opposite order.

So the inverse of "first multiply x by 8 and then subtract 3" is "first add 3 to x and then divide by 8".

$f(x)= \frac{x+ 3}{8}= \frac{1}{8}x+ \frac{3}{8}$