Consider f(x) = [x]|x^3 - 2x^2 - x + 2| in [- | vedclass

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(D) First,factorize the expression inside the absolute value: $x^3 - 2x^2 - x + 2 = x^2(x - 2) - 1(x - 2) = (x^2 - 1)(x - 2) = (x - 1)(x + 1)(x - 2)$.

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$3$

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(D) First,factorize the expression inside the absolute value: $x^3 - 2x^2 - x + 2 = x^2(x - 2) - 1(x - 2) = (x^2 - 1)(x - 2) = (x - 1)(x + 1)(x - 2)$.Thus,$f(x) = [x]|(x - 1)(x + 1)(x - 2)|$.The function $g(x) = |(x - 1)(x + 1)(x - 2)|$ is a continuous polynomial function.The function $h(x) = [x]$ is discontinuous at all integers $x \in \mathbb{Z}$.In the interval $[-\frac{3}{2}, \frac{9}{2}]$,the integers are $\{-1, 0, 1, 2, 3, 4\}$.Let $f(x) = [x] \cdot g(x)$. The product is discontinuous at $x = k$ if $g(k) 
eq 0$ and $[x]$ is discontinuous at $x = k$.Check the values of $g(k)$ for $k \in \{-1, 0, 1, 2, 3, 4\}$:$g(-1) = |(-1-1)(-1+1)(-1-2)| = 0$.$g(0) = |(0-1)(0+1)(0-2)| = |(-1)(1)(-2)| = 2 
eq 0$.$g(1) = |(1-1)(1+1)(1-2)| = 0$.$g(2) = |(2-1)(2+1)(2-2)| = 0$.$g(3) = |(3-1)(3+1)(3-2)| = |(2)(4)(1)| = 8 
eq 0$.$g(4) = |(4-1)(4+1)(4-2)| = |(3)(5)(2)| = 30 
eq 0$.Since $g(k) = 0$ at $x = -1, 1, 2$,the function $f(x)$ is continuous at these points because $\lim_{x 	o k} [x]g(x) = g(k) = 0$.Thus,$f(x)$ is discontinuous only at $x = 0, 3, 4$.The number of points of discontinuity is $3$.

Consider $f(x) = [x]|x^3 - 2x^2 - x + 2|$ in $[-\frac{3}{2}, \frac{9}{2}]$. The number of points where $f(x)$ is discontinuous is (where $[.]$ denotes the greatest integer function).

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