Let f and g be two functions defined by f(x) | vedclass

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(B) First, we simplify $f(x)$ as $f(x) = \begin{cases} x+1, & x < 0 \ 1-x, & 0 \leq x < 1 \ x-1, & x \geq 1 \end{cases}$.Now, we find $(g \circ f)(x

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Continuous everywhere but not differentiable exactly at one point

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(B) First, we simplify $f(x)$ as $f(x) = \begin{cases} x+1, & x Now, we find $(g \circ f)(x) = g(f(x))$.If $f(x) If $f(x) \geq 0$, then $x \geq -1$. In this case, $g(f(x)) = 1$.Thus, $(g \circ f)(x) = \begin{cases} x+2, & x Checking continuity at $x = -1$: $\lim_{x 	o -1^-} (x+2) = 1$ and $\lim_{x 	o -1^+} (1) = 1$. Since the limits are equal to $g(f(-1)) = 1$, the function is continuous everywhere.Checking differentiability at $x = -1$: The left-hand derivative is $\frac{d}{dx}(x+2) = 1$, and the right-hand derivative is $\frac{d}{dx}(1) = 0$. Since $1 
eq 0$, it is not differentiable at $x = -1$.

Let $f$ and $g$ be two functions defined by $f(x) = \begin{cases} x+1, & x < 0 \\ |x-1|, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ 1, & x \geq 0 \end{cases}$. Then $(g \circ f)(x)$ is

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