Maximum vertical distance for concave function.

Question

I need to find the maximum vertical distance between a parabola $f(x)=3-x^2$ and a line $g(x)=x+1$ on the interval of $[-2,1]$. Usually I see people doing $g(x)-f(x)$, in other words subtracting the parabola from the line. But since in this case we have a concave function does it mean I have to do $f(x)-g(x)$? Or is it always "line - parabola"?

Answer 1

$f(x)\geq g(x)$ for all $-2\leq x\leq1$ and

$$f(x)-g(x)=3-x^2-x-1=\frac{9}{4}-\frac{1}{4}-x-x^2=\frac{9}{4}-\left(\frac{1}{2}+x\right)^2\leq\frac{9}{4}.$$ The equality occurs for $x=-\frac{1}{2}\in[-2,1]$,

which says that $\frac{9}{4}$ is the answer.

Answer 2

It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$ If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.