Question

Let f(x)=2/x-3. Find f^-1(x)

Answer 1

Two questions:

Question 1:
`f^(-1)(x)=?` given
`f(x)=(2)/(x)-3`.

Answer 1:
`f^(-1)(x)=(2)/(x+3)`

Question 2:
`f^(-1)(x)=?` given
`f(x)=(2)/(x-3)`.

Answer 2:
`f^(-1)(x)=(2)/(x)+3`

Explanation:

So
`f^(-1)` is used in most classes to represent the inverse function of
`f`.

The inverse when graphed is a reflection through the y=x line. The ordered pairs
`(a,b)` on
`f` implies
`(b,a)` are on
`f^(-1)`.

This means we really just need to swap x and y.

Since we want to write as a function of x we will need to solve for y again.

Question 1:

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

Swap x and y:

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

We want to solve for y.

Add 3 on both sides:

`x+3=(2)/(y)`

Make the left hand side a fraction so we can cross-multiply:

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

Cross multiply:

`y(x+3)=1(2)`

Simplify right hand side:

`y(x+3)=2`

Divide both sides by (x+3):

`y=(2)/(x+3)`

So
`f^(-1)(x)=(2)/(x+3)`.

Question 2:

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

Swap x and y:

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

Make left hand side a fraction so we can cross multiply:

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

Cross multiply:

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

We have to distribute here:

`yx-3x=2`

Add 3x on both sides:

`yx=2+3x`

Divide boht sides by x:

`y=(2+3x)/(x)`

You could probably stop here but you could also simplify a little.

Separate the fraction into two terms since you have 2 terms on top bottom being dividing by x:

`y=(2)/(x)+(3x)/(x)`

Simplify second fraction x/x=1:

`y=(2)/(x)+3`

So
`f^(-1)(x)=(2)/(x)+3`.