Let the function f be defined by f(x) = frac{ | vedclass

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(B) For $x > 0$, $|x| = x$, so $f(x) = \frac{x - x}{x} = 0$.For $x < 0$, $|x| = -x$, so $f(x) = \frac{x - (-x)}{x} = \frac{2x}{x} = 2$.At $x = 0$, $f(

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continuous for all $x$ except at $x = 0$

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(B) For $x > 0$, $|x| = x$, so $f(x) = \frac{x - x}{x} = 0$.For $x At $x = 0$, $f(0) = 2$.Now, check continuity at $x = 0$:$\lim_{x 	o 0^+} f(x) = 0$.$\lim_{x 	o 0^-} f(x) = 2$.Since $\lim_{x 	o 0^+} f(x) 
eq \lim_{x 	o 0^-} f(x)$, the limit does not exist at $x = 0$, so $f$ is discontinuous at $x = 0$.For $x > 0$, $f(x) = 0$ (a constant function), which is continuous.For $x Thus, $f$ is continuous for all $x$ except at $x = 0$.

Let the function $f$ be defined by $f(x) = \frac{x - |x|}{x}$ for $x \neq 0$ and $f(0) = 2$. Then $f$ is:

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