Question

What is the inverse function of `y=3x^2-5`?

Answer 1

`x=sqrt((y+5)/3)`

Explanation:

To find the inverse of a function expressed as
`color(white)("XXX")y=` an expression in `x`
rearrange the expression into the form:
`color(white)("XXX")x=` an expression in `y`

Given
`color(white)("XXX")y=3x^2-5`

Switch sides and add `5` to both sides
`color(white)("XXX")3x^2=y+5`

Divide both sides by `3`
`color(white)("XXX")x^2=(y+5)/3`

Take the square root of both sides
`color(white)("XXX")x= +-sqrt((y+5)/3)`

Since we were asked for an inverse function, I've eliminated the non-primary root (a function can not have two values for a single argument value).

Therefore
`color(white)("XXX")x=sqrt((y+5)/3)`

Answer 2

With the default domain, this function has no inverse since it is not one-to-one, but read on...

Explanation:

Let `f(x) = 3x^2-5`.

The implicit domain of `f` is the set of values for which it is defined.

By convention, since the variable name is `x` rather than `z`, we're talking about `RR` rather than `CC` and `f` is well defined for the whole of `RR`.

However, `f` is not one-to-one in that `f(-x) = f(x)` for all `x in RR`, so `f` has no well-defined inverse on its default domain.

We can try to find an inverse as follows:

Let `y = f(x) = 3x^2-5`.

Adding `5` to both ends we get:

`y+5 = 3x^2`

Dividing both sides by `3` we get:

`x^2 = (y+5)/3`

Hence

`x = +-sqrt((y+5)/3)`

This is not uniquely defined, so does not define a function, unless...

If we restrict the domain of `f` to `[0, oo)` then `f` is one-to-one and has inverse:

`f^(-1)(y) = sqrt((y+5)/3)`

Alternatively, if we restrict the domain of `f` to `(-oo, 0]`, then `f` is one-to-one with inverse:

`f^(-1)(y) = -sqrt((y+5)/3)`

Interestingly, if we define `g(y) = sqrt((y+5)/3)`, then `g(y)` has an inverse function `g^(-1)(x) = f(x)`.

If we define `h(y) = -sqrt((y+5)/3)`, then `h(y)` has an inverse function `h^(-1)(x) = f(x)`.

If we define:

`k(y) = { (sqrt((y+5)/3), "if " y " is rational"), (-sqrt((y+5)/3), "if " y " is irrational") :}`

then `k(y)` has an inverse function `k^(-1)(x) = f(x)`

The implicit domain for `g(y), h(y) and k(y)` is `[-5, oo)` since we require `y >= -5` in order that the square root has a Real value.