Let f(x)=[2x^2+1] and g(x)=cases2x-3, & x

Question

Let $f(x)=[2x^2+1]$ and $g(x)=\begin{cases}2x-3, & x<0\ 2x+3, & x\ge0\end{cases}$, where $[t]$ is the greatest integer $\le t$. Then, in the open interval $(-1,1)$, the number of points where $f\circ g$ is discontinuous is equal to\_\_\_\_

Answer: 62

Accepted Answer

$$f(g(x))=[2g(x)^2]+1=\begin{cases}[2(2x-3)^2]+1, & x<0\ [2(2x+3)^2]+1, & x\ge0\end{cases}$$

$f\circ g$ is discontinuous wherever $2(2x-3)^2$ or $2(2x+3)^2$ passes through an integer value (a jump of the greatest-integer function), for $x\ne0$. Counting all such jump points within $(-1,1)$ across the two branches gives 62 points of discontinuity.

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