Question

Is this how you solve this (finding the inverse)

Answer 1

There are a couple ways to go about this.

By definition of inverse function, if
`j^(-1)` exists, then

`j\left(j^(-1)(x)\right) = x`

All this means is that if you provide some input
`x`, plug it into
`j^(-1)(x)`, then plug that into
`j(x)`, you end up recovering the original input
`x` that you started with.

We can also do this in the opposite direction, so

`j^(-1)(j(x)) = x`

So now if
`j(x)` is invertible, then

`j(x) = -\frac23 x - 9 \implies j\left(j^(-1)(x)\right) = -\frac23 j^(-1)(x) - 9 = x`

Solve for
`j^(-1)`.

`-\frac23 j^(-1)(x) - 9 = x`

`\frac23 j^(-1)(x) = -x - 9`

`j^(-1)(x) = -\frac32 (x+9)`

`\boxed{j^(-1)(x) = -\frac32 x - \frac{27}2}`

Another way to do it is by algebraically substituting
`x` within the definition of
`j(x)`

Starting with the given definition

`j(x) = -\frac23 x - 9`

we replace
`x` with some new expression in a new variable
`y` that makes the right side reduce to just
`y`. To demonstrate, if we multiply
`-\frac23 x` by
`-\frac32`, we end up with a coefficient of 1. So let
`x=-\frac32 y`. Then

`j\left(-\frac32 y\right) = -\frac23 \left(-\frac32 y\right) - 9 = y - 9`

As you can see, this doesn't completely reduce to
`y`, so we're not done. To adjust for this, we just add 9 to
`y`, so our ultimate substitution is
`x=-\frac32(y+9)`. Then

`j\left(-\frac32 (y+9)\right) = -\frac23 \left(-\frac32 (y+9)\right) - 9 = (y+9) - 9 = y`

Now applying
`j^(-1)` to both sides,

`j^(-1)\left(j\left(-\frac32 (y+9)\right)\right) = j^(-1)(y)`

`\implies \boxed{j^(-1)(y) = -\frac32 y - \frac{27}2}`

(we can freely swap out
`y` for
`x` at this point)

Yet another way, since the function here is quite simple, we can just think about the action of
`j` on a given number
`x` - multiply it by -2/3, and subtract 9.

`x \mapsto -\frac23 x \mapsto -\frac23x - 9 = j(x)`

The inverse operation undoes this, so given some number
`j(x)`, we recover
`x` by adding the additive inverse of -9, or 9, then multiplying the whole result by the multiplicative inverse of -2/3, or -3/2.

`j(x) \mapsto j(x) + 9 \mapsto -\frac32(j(x) + 9) \mapsto -\frac32 j(x) - \frac{27}2 = x`

You seem to be using some version of this method. You made the mistake of thinking the "multiply by -3/2" step applies only to
`j(x)`, and not
`j(x) + 9`.