What is the largest interval containing $0$ so that $f(x)=(x+1)^2$ has an inverse? What is the largest possible interval containing $0$ such that $f$ has an inverse on that interval, where $f(x)=(x+1)^2$?

I have tried to find the inverse and the derivative.

$f(x)$ is $0$ when $x=-1$. 

$f$ derivative is $0$ when $x=-2$; is this the local minimum?

$f$ inverse is $0$  when $x=1$; is this the local maximum?
For a function to admit a restricted inverse, one has to check injectivity (if a function is not injective in $(a,b)$ it for sure does not admit an inverse function in that interval).

Now, check that the function you gave is injective in the interval $(-\infty,-1]$ (for example check that the sign of the derivative there is always negative, therefore the function is decreasing, hence injective). Analogously, it is injective in $[-1,\infty)$ (change a bit the previous argument to conclude so).

Then, if I had to choose the “largest” interval containing zero for which f admits an inverse I’d go with $[-1,\infty)$, which geometrically corresponds to the right “half” of the parabola.
This is a parabola centered at $(x,y)=(-1,0)$. Notice that both branches are separately invertible. Consider the interval $[-1,\infty)$. This contains $0$ and $f$ is invertible. For any other interval, say, $[a,\infty)$ with $a&lt;-1,$ the function is not one to-one.