Let f(x) = [2x^2 + 1] and g(x) = begin{cases} | vedclass

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(A) We are given $f(x) = [2x^2 + 1] = [2x^2] + 1$. Then $f(g(x)) = [2(g(x))^2] + 1$. Case $1$: $x < 0$,$g(x) = 2x - 3$. $f(g(x)) = [2(2x - 3)^2] + 1$.

Vedclass · Accepted Answer

$62$

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(A) We are given $f(x) = [2x^2 + 1] = [2x^2] + 1$. Then $f(g(x)) = [2(g(x))^2] + 1$. Case $1$: $x $f(g(x)) = [2(2x - 3)^2] + 1$. For $x \in (-1, 0)$,$2x - 3 \in (-5, -3)$. Thus,$(2x - 3)^2 \in (9, 25)$,so $2(2x - 3)^2 \in (18, 50)$. The function $[2(2x - 3)^2] + 1$ is discontinuous when $2(2x - 3)^2$ is an integer. In the interval $(-1, 0)$,$2x - 3$ ranges from $-5$ to $-3$. The values of $2(2x - 3)^2$ range from $18$ to $50$. The integers in $(18, 50)$ are $19, 20, \dots, 49$,which gives $49 - 19 + 1 = 31$ points. Case $2$: $x \geq 0$,$g(x) = 2x + 3$. $f(g(x)) = [2(2x + 3)^2] + 1$. For $x \in [0, 1)$,$2x + 3 \in [3, 5)$. Thus,$(2x + 3)^2 \in [9, 25)$,so $2(2x + 3)^2 \in [18, 50)$. The integers in $[18, 50)$ are $18, 19, \dots, 49$,which gives $49 - 18 + 1 = 32$ points. However,we must check the point $x = 0$. At $x = 0$,$f(g(0)) = f(3) = [2(3)^2 + 1] = [19] = 19$. $lim_{x 	o 0^-} f(g(x)) = [2(-3)^2] + 1 = [18] + 1 = 19$. $lim_{x 	o 0^+} f(g(x)) = [2(3)^2] + 1 = [18] + 1 = 19$. Since the limits are equal,$x = 0$ is not a point of discontinuity. Total points = $31 + 32 - 1 (	ext{for } x=0 	ext{ overlap}) = 62$. Wait,the interval is $(-1, 1)$,so we exclude the endpoints. For $x For $x > 0$,$2(2x+3)^2$ takes values in $(18, 50)$. Integers are $19, \dots, 49$ ($31$ points). Total points = $31 + 31 = 62$.

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

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