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

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#

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.