Maximum minimum of $x^2$ on the open interval $-1<x<1$

Question

The question goes like this:

Does the function $f(x)=x^2$ have a maximum on the open interval $(-1,1)$ ? And a minimum? Explain

Not exactly sure how to give a proper answer really. For me it makes sense that it has a minimum but not a maximum since the interval is open $(1,1)$ instead of closed $[1,1]$. Still, I really do not know how to explain it in mathematical language based upon some mathematical principle...does it also have to do with the Intermediate Value Theorem?

Any suggestion appreciated.

Answer 1

From $f\bigl((-1,1)\bigr)=[0,1)$ follows that $f$ has a minimum but no maximum on $(-1,1)$.

Edit: And yes, it follows from the IVT as it states that the image of an interval under a continuous function itself is an interval.

Answer 2

the Minimum Min(0,0) lies in the interval $(-1,1)$ and annsupremum can be found by $\lim_{x -> \pm 1}x^2$