If a and b (a b) are points of discontinuit | vedclass

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(C) To find the points of discontinuity,we check the limits at the transition points $x = 0, 1, 2, 3$. At $x = 0$: $f(0) = 3-2(0)^2 = 3$. $\lim_{x 	o

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$5$

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(C) To find the points of discontinuity,we check the limits at the transition points $x = 0, 1, 2, 3$. At $x = 0$: $f(0) = 3-2(0)^2 = 3$. $\lim_{x 	o 0^-} f(x) = 3$ and $\lim_{x 	o 0^+} f(x) = 2(0)+3 = 3$. Continuous. At $x = 1$: $\lim_{x 	o 1^-} f(x) = 2(1)+3 = 5$. $\lim_{x 	o 1^+} f(x) = 2(1)^2-3(1) = -1$. Since $5 
eq -1$,$x = 1$ is a point of discontinuity. At $x = 2$: $\lim_{x 	o 2^-} f(x) = 2(2)^2-3(2) = 8-6 = 2$. $\lim_{x 	o 2^+} f(x) = 2(2)-3 = 1$. Since $2 
eq 1$,$x = 2$ is a point of discontinuity. At $x = 3$: $\lim_{x 	o 3^-} f(x) = 2(3)-3 = 3$. $\lim_{x 	o 3^+} f(x) = |3| = 3$. Continuous. Thus,the points of discontinuity are $a = 2$ and $b = 1$ (given $a > b$). Therefore,$3a - b = 3(2) - 1 = 6 - 1 = 5$.

If $a$ and $b$ $(a > b)$ are points of discontinuity of the function $f(x) = \begin{cases} 3-2x^2, & \text{for } x \leq 0 \\ 2x+3, & \text{for } 0 < x \leq 1 \\ 2x^2-3x, & \text{for } 1 < x < 2 \\ 2x-3, & \text{for } 2 \leq x < 3 \\ |x|, & \text{for } x \geq 3 \end{cases}$,then $3a-b = $

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