Let f(x) = begin{cases} x+1, & -1 leq x leq 0 | vedclass

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(C) The function is defined as $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \ -x, & 0 < x \leq 1 \end{cases}$.Check continuity at $x=0$:Left-hand li

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$f(x)$ is discontinuous in $[-1,1]$ but still has the maximum and minimum value

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(C) The function is defined as $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \ -x, & 0 Check continuity at $x=0$:Left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (x+1) = 1$.Right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (-x) = 0$.Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{x 	o 0^+} f(x)$, the function is discontinuous at $x=0$.Now, analyze the range of $f(x)$ on $[-1, 1]$:For $x \in [-1, 0]$, $f(x) = x+1$. The range is $[0, 1]$.For $x \in (0, 1]$, $f(x) = -x$. The range is $[-1, 0)$.Combining these, the range of $f(x)$ is $[-1, 0) \cup [0, 1] = [-1, 1]$.The maximum value is $1$ (attained at $x=0$) and the minimum value is $-1$ (attained at $x=1$).Thus, $f(x)$ is discontinuous in $[-1, 1]$ but still has a maximum and minimum value.

Let $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \\ -x, & 0 < x \leq 1 \end{cases}$. Which of the following statements is true?

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