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

Question

Lincoln Anderson · Accepted Answer

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

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#

Lincoln Anderson · Answer

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

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.

Theodore Amidon · Answer

undefined The inverse function of $ f(x) = (x + 1)^2 $ is $ f^{-1}(x) = \sqrt{x} - 1 $.