Question 1187396
The function {{{f(x)=(x+1)^2}}} is a quadratic function, meaning a polynomial function of degree 2.
Those functions have graphs like this {{{graph(110,100,2,7.5,1.5,6.5,(x-5)^2+2)}}} or the same curve upside down.
They are symmetrical with a vertex that is a maximum or minimum, separating a decreasing branch from an increasing branch.
You realize that for {{{x=-1}}} , {{{x+1=0}}} and {{{f(-1)=0^2=0}}} , but for any other number {{{f(x)>0}}} .
The graph decreases from any value of {{{x<-1}}} , and increases for {{{x>=-1}}} .
The largest domain on which f is one-to-one and non-decreasing is [-1,infinity), or {{{x>=-1}}} .
The range is [0,infinity), because {{{f(-1)=0}}}, and for {{{x>=-1}}} it increases without bounds.
To find the inverse we swap {{{y}}} and {{{x}}} in the function defined as {{{y=(x+1)^2}}} only when {{{x>=-1}}}<-->{{{x+1>=0}}} , and solve for {{{y}}}
we get {{{x=(y+1)^2}}} for {{{y+1>=0}}} , whose solution is {{{sqrt(x)=y+1}}} <--> {{{y=sqrt(x)-1}}} .
The inverse function is {{{f^-1}}}{{{(x)=sqrt(x)-1}}} for {{{X>=0}}} . 
{{{f^-1}}}{{{(x)=sqrt(x)-1}}} has domain [0,infinity) or {{{x>=0}}} and range [-1,infinity) or {{{y>=-1}}}
The domain of an inverse function is the range of the domain-restricted function, and the range of the inverse is the restricted dominion.