Find the absolute minimum value of the function $y=3x^2-4$ in the interval $[-1,5]$.

1. Find critical points: $y^{\prime }=6x$. $6x=0⟹x=0$.2. Check if $0\in [-1,5]$. Yes.3. Evaluate function at critical point and endpoints:$y(0)=3(0{)}^2-4=-4$$y(-1)=3(-1{)}^2-4=3-4=-1$$y(5)=3(5{)}^2-4=75-4=71$ The absolute minimum is the smallest of $\{-4,-1,71\}$, which is $-4$.