The number of points where the function f(x) | vedclass

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(A) To find the points of discontinuity,we check the transition points and the points where the internal functions are discontinuous.$1$. For $x \leq

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$9$

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(A) To find the points of discontinuity,we check the transition points and the points where the internal functions are discontinuous.$1$. For $x \leq -1$,$f(x) = |2x^2 - 3x - 7|$. This is a continuous function.At $x = -1$,$f(-1) = |2(-1)^2 - 3(-1) - 7| = |2 + 3 - 7| = |-2| = 2$.$2$. For $-1 $4x^2 = k + 1 \implies x^2 = \frac{k+1}{4} \implies x = \pm \frac{\sqrt{k+1}}{2}$.For $-1 - If $k = -1$,$x^2 = 0 \implies x = 0$.- If $k = 0$,$x^2 = 1/4 \implies x = \pm 1/2$.- If $k = 1$,$x^2 = 2/4 = 1/2 \implies x = \pm 1/\sqrt{2}$.- If $k = 2$,$x^2 = 3/4 \implies x = \pm \sqrt{3}/2$.These are $7$ points: $0, 1/2, -1/2, 1/\sqrt{2}, -1/\sqrt{2}, \sqrt{3}/2, -\sqrt{3}/2$.$3$. At $x = 1$,$f(1) = |1+1| + |1-2| = 2 + 1 = 3$. From the left,$\lim_{x 	o 1^-} f(x) = [4(1)^2 - 1] = [3] = 3$ (if we consider the limit as $x$ approaches $1$ from the left,$4x^2-1$ approaches $3$,so $[4x^2-1]$ approaches $2$). Since $f(1) = 3$ and $\lim_{x 	o 1^-} f(x) = 2$,it is discontinuous at $x = 1$.$4$. At $x = -1$,$\lim_{x 	o -1^+} f(x) = [4(-1)^2 - 1] = [3] = 3$. Since $f(-1) = 2$,it is discontinuous at $x = -1$.Total points of discontinuity: $7$ points from the interval $(-1, 1)$ plus the points $x = -1$ and $x = 1$. However,checking the graph and the function definition,the points of discontinuity are $x \in \{-1, -\sqrt{3}/2, -1/\sqrt{2}, -1/2, 0, 1/2, 1/\sqrt{2}, \sqrt{3}/2, 1\}$. Counting these,we get $9$ points.

The number of points where the function $f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \leq -1 \\ [4x^2 - 1] & \text{if } -1 < x < 1 \\ |x+1| + |x-2| & \text{if } x \geq 1 \end{cases}$ is discontinuous,where $[t]$ denotes the greatest integer $\leq t$,is:

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