Question

FInd the inverse of `f(x)=1+x/3-x` ?

Having some trouble here, i'd appreciate it if I got some help.

Answer 1

`f^-1(x)=-3/2(y-1)`

Explanation:

To find the `f^-1(x)` of a function `f(x)=y=....x+....`, we make `x` in terms of `y`.

In here,

`f(x)=1+x/3-x`

`f(x)=1+1/3x-x`

`f(x)=1-2/3x`

Now, just put `y=f(x)`

`y=1-2/3x`

`-2/3x=y-1`

`x=-3/2(y-1)`

Therefore,

`f^-1(x)=-3/2(y-1)`

Answer 2

Given `f(x) =(1+x)/(3-x); x!=3`

Begin by substituting `x = f^-1(x)` into `f(x)`:

`f(f^-1(x)) =(1+f^-1(x))/(3-f^-1(x))`

Use the property `f(f^-1(x)) = x` to make the left side become x:

`x =(1+f^-1(x))/(3-f^-1(x))`

Solve for `f^-1(x)`:

`x(3-f^-1(x))=1+f^-1(x))`

`3x-xf^-1(x)=1+f^-1(x))`

`3x-1-xf^-1(x)=f^-1(x))`

`3x-1=xf^-1(x)+f^-1(x))`

`3x-1=(x+1)f^-1(x))`

`f^-1(x) = (3x-1)/(x+1); x !=-1`

To verify that the above is truly the inverse, you should show that `f(f^-1(x)) = x` and `f^-1(f(x)) = x`