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

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- Oct 27th 2009, 09:12 AMOasis1993Inverse 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 - Oct 27th 2009, 09:25 AMramiee2010
let f(x)= ax+b =y (say)

then ax+b =y $\displaystyle \quad \Rightarrow \quad x= \frac{(y-b)}{a} $

$\displaystyle or\ x=\frac{1}{a}y-\frac{b}{a} $

$\displaystyle \therefore f^{-1} (y)=\frac{1}{a}y-\frac{b}{a} \quad \Rightarrow \quad (\because f(x)=y , taking inverse both side )$

$\displaystyle or\ \color {blue} f^{-1} (x)=\frac{1}{a}x-\frac{b}{a}$................(1)

but given $\displaystyle f^{-1}(x)=8x-3$................(2)

on comparison eq(1) and (2)

$\displaystyle \frac{1}{a}=8 \quad \Rightarrow a= \frac{1}{8}$

and

$\displaystyle \frac{b}{a}=3 \quad \Rightarrow b=3a={3\over 8} $ - Oct 27th 2009, 11:30 AMOasis1993
Thank you very much!

- Oct 28th 2009, 07:52 AMHallsofIvy
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".

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