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We are given the piecewise function:
![\displaystyle \large{f(x) = \begin{cases} {x}^(3) \: \: (x < 0) \\ \sqrt[3]{x} \: \: (x > 0) \end{cases}}](https://img.qammunity.org/2022/formulas/mathematics/high-school/1dmtra3vxhhih7r5kgtci3skarghjdph6h.png)
To evaluate the value of function at x = -8 and x = 8, we know that -8 is less than 0 and 8 is greater than 0.
Therefore, if we want to evaluate the value of function at x = -8; we use the x^3 since it's given x < 0 for the function and for x = 8; we use the cube root of x since it's given x > 0.
Evaluate x = -8
From the piecewise function:
![\displaystyle \large{f(x) = \begin{cases} {x}^(3) \: \: (x < 0) \\ \sqrt[3]{x} \: \: (x > 0) \end{cases}}](https://img.qammunity.org/2022/formulas/mathematics/high-school/1dmtra3vxhhih7r5kgtci3skarghjdph6h.png)
Since -8 is less than 0, we use x^3.
![\displaystyle \large{f( - 8) = {( - 8)}^(3) } \\ \displaystyle \large{f( - 8) = - 512}](https://img.qammunity.org/2022/formulas/mathematics/high-school/hd75pdwlnpiw4x0lzjrnmgi623zkqlx34a.png)
Evaluate x = 8
From the piecewise function, since 8 is greater than 0, we use the cube root of x.
![\displaystyle \large{f(8) = \sqrt[3]{8} }](https://img.qammunity.org/2022/formulas/mathematics/high-school/rx2in5xhapo8fdgx59t7p0mycaedip8dlg.png)
To evaluate the cube root, first we prime-factor the number.
![\displaystyle \large{f(8) = \sqrt[3]{2 \cdot 2 \cdot 2} }](https://img.qammunity.org/2022/formulas/mathematics/high-school/lfust18wrbiewvd9y9onp9h2xru2jjd73e.png)
Since it's a cube root, we pull three 2's out of the cube root and write only one 2.
![\displaystyle \large{f(8) =2}](https://img.qammunity.org/2022/formulas/mathematics/high-school/hmu1t70hcog7t1y7wk18a404v19xhs6w2k.png)
Answer
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