Question

Find the min value of function $f\left(x\right)=\{\begin{array}{c}{x}^2+1x<0\\ xx\ge 0\end{array}$

Answer 1

In the function the whole negative numbers turn out to be positive and will be $>1$ since the function is $x^2+1$

And for $x\ge 0$ it is strictly increasing function

So the min value is at $x=0$ where $f\left(x\right)=0$