If the function f:[-1,1] rightarrow R is defi | vedclass

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(D) The function $f(x)$ is defined on the interval $[-1, 1]$.For $x \in [-1, 0)$,$f(x) = 2^x + 1$. As $x 	o 0^-$,$f(x) 	o 2^0 + 1 = 2$. Since $2^x$

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neither maximum nor minimum

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(D) The function $f(x)$ is defined on the interval $[-1, 1]$.For $x \in [-1, 0)$,$f(x) = 2^x + 1$. As $x 	o 0^-$,$f(x) 	o 2^0 + 1 = 2$. Since $2^x$ is strictly increasing,$f(x)$ increases from $f(-1) = 2^{-1} + 1 = 1.5$ to $2$.At $x = 0$,$f(0) = 1$.For $x \in (0, 1]$,$f(x) = 2^x - 1$. As $x 	o 0^+$,$f(x) 	o 2^0 - 1 = 0$. Since $2^x$ is strictly increasing,$f(x)$ increases from $0$ to $f(1) = 2^1 - 1 = 1$.The range of the function is $[1.5, 2) \cup \{1\} \cup (0, 1]$.Combining these,the range is $(0, 1] \cup [1.5, 2)$.Since the range is not a closed interval and the function values do not attain a supremum or infimum within the domain (the values approach $2$ but never reach it,and approach $0$ but never reach it),the function has neither a maximum nor a minimum value on $[-1, 1]$.

If the function $f:[-1,1] \rightarrow R$ is defined by $f(x) = \begin{cases} 2^x+1, & \text{for } x \in [-1,0) \\ 1, & \text{for } x=0 \\ 2^x-1, & \text{for } x \in (0,1] \end{cases}$,then in $[-1,1]$,$f(x)$ has

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