Let f(x)=√2,x, x^2.

Question

Let $f(x)=\min\{\sqrt2\,x,\ x^2\}$. Let $S=\{x\in(-2,2): g(x)=|x|[x^2]	ext{ is discontinuous}\}$, where $[\cdot]$ denotes the greatest integer function. Then $\displaystyle\sum_{x\in S}f(x)$ equals

Accepted Answer

$g(x)=|x|[x^2]$ is discontinuous wherever $x^2$ passes through an integer value (a jump point of the greatest integer function), within $(-2,2)$ i.e. $x^2<4$: these are $x^2=1,2,3$, giving
$$S=\{-1,\,1,\,-\sqrt2,\,\sqrt2,\,-\sqrt3,\,\sqrt3\}$$

Now compute $f(x)=\min\{\sqrt2x,x^2\}$ at each point:

- $f(-1)=\min(-\sqrt2,1)=-\sqrt2$
- $f(1)=\min(\sqrt2,1)=1$
- $f(-\sqrt2)=\min(-2,2)=-2$
- $f(\sqrt2)=\min(2,2)=2$
- $f(-\sqrt3)=\min(-\sqrt6,3)=-\sqrt6$
- $f(\sqrt3)=\min(\sqrt6,3)=\sqrt6$

$$\sum_{x\in S}f(x)=(-\sqrt2+1)+(-2+2)+(-\sqrt6+\sqrt6)=1-\sqrt2$$

Solutions · 1