f(x) = begin{cases} 0, & x = 0 x - 3, & x | vedclass

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(B) The given function is $f(x) = \begin{cases} 0, & x = 0 \ x - 3, & x > 0 \end{cases}$.First,let us check the continuity at $x = 0$.$f(0) = 0$.$\li

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strictly increasing when $x > 0$

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(B) The given function is $f(x) = \begin{cases} 0, & x = 0 \ x - 3, & x > 0 \end{cases}$.First,let us check the continuity at $x = 0$.$f(0) = 0$.$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x - 3) = -3$.Since $f(0) 
eq \lim_{x 	o 0^+} f(x)$,the function is discontinuous at $x = 0$.Now,consider the interval $x > 0$. For any $x_1, x_2$ such that $0  0$.However,for the function to be increasing on the interval $[0, \infty)$,it must be continuous on $[0, \infty)$. Since it is discontinuous at $x = 0$,it is not increasing on $[0, \infty)$.Thus,the function is strictly increasing when $x > 0$.

$f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$. The function $f(x)$ is

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