If f(x) = begin{cases} -x^2, & text{when } x | vedclass

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(B) To check continuity at $x = 0$:$\lim_{x 	o 0^-} f(x) = -(0)^2 = 0$$\lim_{x 	o 0^+} f(x) = 5(0) - 4 = -4$Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{

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$f(x)$ is continuous at $x = 2$

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(B) To check continuity at $x = 0$:$\lim_{x 	o 0^-} f(x) = -(0)^2 = 0$$\lim_{x 	o 0^+} f(x) = 5(0) - 4 = -4$Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{x 	o 0^+} f(x)$,$f(x)$ is discontinuous at $x = 0$.To check continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = 5(1) - 4 = 1$$\lim_{x 	o 1^+} f(x) = 4(1)^2 - 3(1) = 1$$f(1) = 5(1) - 4 = 1$Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$,$f(x)$ is continuous at $x = 1$.To check continuity at $x = 2$:$\lim_{x 	o 2^-} f(x) = 4(2)^2 - 3(2) = 16 - 6 = 10$$\lim_{x 	o 2^+} f(x) = 3(2) + 4 = 10$$f(2) = 3(2) + 4 = 10$Since $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2)$,$f(x)$ is continuous at $x = 2$.Thus,option $B$ is correct.

If $f(x) = \begin{cases} -x^2, & \text{when } x \le 0 \\ 5x - 4, & \text{when } 0 < x \le 1 \\ 4x^2 - 3x, & \text{when } 1 < x < 2 \\ 3x + 4, & \text{when } x \ge 2 \end{cases}$,then:

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