If f:[-3,2] rightarrow [0, sqrt[3]{x}] is an | vedclass

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(C) Given the function $f:[-3,2] ightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \ n^{2/3}, & -1 <

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$4$

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(C) Given the function $f:[-3,2] ightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \ n^{2/3}, & -1 For the function to be onto,the range of $f(n)$ must be equal to the codomain $[0, \sqrt[3]{x}]$.First,evaluate the function at the boundaries and critical points:For $-3 \leq n \leq -1$,$f(n)$ increases from $f(-3) = 2 + \sqrt[3]{-3} \approx 0.55$ to $f(-1) = 2 + (-1)^{1/3} = 2 - 1 = 1$.For $-1 Since the function is onto,the maximum value of the function must be the upper bound of the codomain.The maximum value is $\max(f(-1), f(2)) = \max(1, \sqrt[3]{4}) = \sqrt[3]{4}$.Thus,$\sqrt[3]{x} = \sqrt[3]{4}$,which implies $x = 4$.

If $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ is an onto function defined by $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$,then $x=$

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