Mathematics

Mathematics

Problem

Let f(x)={x2−4x,x≤1,6−2x,x>1.f(x)=\begin{cases}x^{2}-4x,&x\le 1,\\6-2x,&x>1.\end{cases} For the accumulation function F(x)=∫−2xf(t) dtF(x)=\int_{-2}^{x} f(t)\,dt, state the open interval(s) on which FF is concave down.
Intuitive explanation

The key is recognizing that concavity of FF depends on the sign of F′′(x)F''(x), which equals f′(x)f'(x) by the Fundamental Theorem of Calculus. You need to find where f′(x)<0f'(x) < 0 by differentiating each piece of ff, then carefully check the boundary at x=1x=1 where the formula changes. The concavity flips wherever the derivative of ff changes from positive to negative.

Intuitive explanation

The key is recognizing that concavity of FF depends on the sign of F′′(x)F''(x), which equals f′(x)f'(x) by the Fundamental Theorem of Calculus. You need to find where f′(x)<0f'(x) < 0 by differentiating each piece of ff, then carefully check the boundary at x=1x=1 where the formula changes. The concavity flips wherever the derivative of ff changes from positive to negative.

Browse the problem library