Mathematics

Problem

The function

f

is continuous on

0\le x\le 8

and differentiable on

(0,4)

and

(4,8)

. It is strictly increasing on

(0,4)

and strictly decreasing on

(4,8)

. Let

F(x)=\displaystyle \int_{2}^{x} f(t)\,dt

. On which interval must

F

be concave down? a)

(0,4)

(4,8)

(2,6)

(2,4)

Intuitive explanation

The key is recognizing that $F'(x) = f(x)$ by the Fundamental Theorem of Calculus, so $F$ is concave down wherever $f$ is decreasing. Since $f$ is strictly decreasing on $(4,8)$ , that's where $F''(x) = f'(x) < 0$ . The challenge is determining whether $F$ can be concave down on $(0,4)$ — but $f$ is strictly increasing there, so $F$ is concave up instead.

Intuitive explanation

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