Question

What’s the inverse function

Answer 1

`\laege\boxed{f^(-1)(x)=-2x+6}`

Explanation:

`f(x)=3-(1)/(2)x\to y=3-(1)/(2)x\\\\\text{Exchange x to y and vice versa:}\\\\x=3-(1)/(2)y\\\\\text{Solve for}\ y:\\\\3-(1)/(2)y=x\qquad\text{subtract 3 from both sides}\\\\-(1)/(2)y=x-3\qquad\text{multiply both sides by (-2)}\\\\\left(-2\!\!\!\!\diagup^1\right)\cdot\left(-(1)/(2\!\!\!\!\diagup_1)y\right)=-2x-3(-2)\\\\y=-2x+6`

Answer 2

`f^(-1)(x)=-2x+6`.

Explanation:

`y=f(x)`

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

The biggest thing about finding the inverse is swapping x and y. The inverse comes from switching all the points on the graph of the original. So a point (x,y) on the original becomes (y,x) on the original's inverse.

Sway x and y in:

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

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

Now we want to remake y the subject (that is solve for y):

Subtract 3 on both sides:

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

Multiply both sides by -2:

`-2(x-3)=y`

We could leave as this or we could distribute:

`-2x+6=y`

The inverse equations is
`y=-2x+6`.

Now some people rename this
`f^(-1)` or just call it another name like
`g`.

`f^(-1)(x)=-2x+6`.

Let's verify this is the inverse.

If they are inverses then you will have that:

`f(f^(-1)(x))=x \text{ and } f^(-1)(f(x))=x`

Let's try the first:

`f(f^(-1)(x))`

`f(-2x+6)` (Replace inverse f with -2x+6 since we had
`f^(-1))(x)=-2x+6`

`3-(1)/(2)(-2x+6)` (Replace old output, x, in f with new input, -2x+6)

`3+x-3` (I distributed)

`x`

Bingo!

Let's try the other way.

`f^(-1)(f(x))`

`f^(-1)(3-(1)/(2)x)` (Replace f(x) with 3-(1/2)x since
`f(x)=3-(1)/(2)x`)

`-2(3-(1)/(2)x)+6` (Replace old input, x, in -2x+6 with 3-(1/2)x since
`f(x)=3-(1)/(2)x`)

`-6+x+6` (I distributed)

`x`

So both ways we got x.

We have confirmed what we found is the inverse of the original function.