Question

What is the inverse function of `f(x)=(x+1)^2`?

Answer 1

The function is not one-to-one. It does not have an inverse function. However,

Explanation:

If we restrict the domain to `[-1,oo)`, then this new function has an inverse:

`g(x)=(x+1)^2` with `x >= -1`

`y=(x+1)^2`
if and only if
`x+1 = sqrty` (We do not need `+-sqrty` because of the restriction on `x`.)

`x=sqrty-1`

`g^-1(x) = -sqrtx-1`

If we restrict the domain to `(-oo,-1]`, we get

`h(x) = (x+1)^2` with `x <= -1`

`y=(x+1)^2`
if and only if
`x+1 = - sqrty` (We need only `-sqrty` because of the restriction on `x`.)

`x= - sqrty-1`

`h^-1(x) = sqrtx-1`

Answer 2

The function is not one-one and therefore there is not a unique inverse function.

Explanation:

First note that the slope (first derivative) is

`f'(x) = 2(x + 1)(1) = 2x + 2` (chain rule).

Also note that

`0 < 2 x + 2`

requires

`-2 < 2x`

that is

`x > -1`

That is, the slope is negative for values of `x` less than `-1` and positive for values of `x` greater than `-1` (and has a stationary point at which the slope is zero at `x = -1`).

That is, the function is not a strictly rising or a strictly descending one.

That is, it is not bijective (one-one).

That is, there is no single inverse function if the domain is taken as the reals.

The function is strictly descending on the open interval `(-oo, -1)` and strictly rising on the open interval `(-1, oo)` so there will be inverse functions on these domains.

Setting `y = (x + 1)^2`

This implies

`(x + 1) = +-sqrt(y)`

which in turn implies

`x = sqrt(y) - 1`

or

`x = -sqrt(y) - 1`

Considering `x = sqrt(y) - 1`

Denoting the required inverse function by `g(x)`, this may be rewritten as

`g(x) = sqrt(x) - 1`

For `g(x)` to be a real function, the domain must be restricted to values of `x` equal to or greater than `0`.

Considering `x = -sqrt(y) - 1`

Denoting the required inverse function by `h(x)`, this may be rewritten as

`h(x) = - sqrt(x) - 1`

For `h(x)` to be a real function, the domain must be restricted to values of `x` equal to or greater than `0`.

Plotting both of these inverse functions on the same graph will yield a parabola "on its side", with apex at `(0, -1)`. That is, values of `x > 0` will correspond to two values (one associated with `g(x)` and the other associated with `h(x)`, reflecting the fact that `f(x)` is not a one-one function.