Respuesta :

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[tex]x^2=x+2\implies x^2-x-2=(x-2)(x+1)=0\implies x=-1,x=2[/tex]

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

[tex]d(x)=|(x+2)-x^2|=2+x-x^2[/tex]

Differentiating, we have

[tex]d'(x)=1-2x[/tex]

which has one critical point at

[tex]1-2x=0\implies x=\dfrac12[/tex]

At this point, we have

[tex]d\left(\dfrac12\right)=\dfrac94[/tex]

as the maximum vertical distance.

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