inverse of one-to-one function

[milanna]

hello all

I need to find the inverse of the one-to-one function

f(x)= 5/3x-8

Would that be f(x)= 3x-8/5 ?

thanks

 

[stapel]

Your formatting is ambiguous. Do you mean:


. . . . .$\displaystyle \large{f(x)\,=\,\frac{5}{3x}\,-\,8}$


...as typed, or:


. . . . .$\displaystyle \large{f(x)\,=\,\frac{5}{3x\,-\,8}}$


...or something else? (Similar question for your inverse.)

Thank you.

Eliz.

 

[milanna]

The problem is actually formatted as

. . . . .$\displaystyle \large{f(x)\,=\,\frac{5}{3x\,-\,8}}$


and the inverse I got was

. . . . .$\displaystyle \large{f(x)\,=\,\frac{3x\,-\,8}{5}}$

Sorry for the confusion. Slowly getting better at LaTeX.

 

[stapel]

The inverse function is only very rarely the reciprocal.

The usual process for finding an inverse is as follows:

. . . . .1) Rename "f(x)" as "y".

. . . . .2) Solve for "x=".

. . . . .3) Switch "x" and "y", so you now have "y=".

. . . . .4) Rename the new "y" as "f^-1(x)"
. . . . . . .(if the inverse happens to be a function, that is).

Try following this procedure with your example. In step (2), you'll need to cross-multiply, and then solve for "x=".

If you get stuck, please reply showing your steps. Thank you.

Eliz.

 

[milanna]

My result came out to be

f^-1(x) = (5/3y) + (8/3)

Is that correct?

 

[stapel]

It looks like you skipped step (3), going straight from (2) to (4). Switch the variables, and you'll be fine.

Eliz.