Question

what is the maximum vertical distance between the line y=x+2 and parabola y=x^2 for -1<=x<=2

Answer 1

$x^2=x+2\implies x^2-x-2=(x-2)(x+1)=0\implies x=-1,x=2$

are the intersection points of the line and parabola. $x^2$ is convex, which guarantees that $x+2\ge x^2$ over this interval. This means the vertical distance between the two functions is

$d(x)=|(x+2)-x^2|=2+x-x^2$

Differentiating, we have

$d'(x)=1-2x$

which has one critical point at

$1-2x=0\implies x=\dfrac12$

At this point, we have

$d\left(\dfrac12\right)=\dfrac94$

as the maximum vertical distance.