Hey there! :)
We are given the piecewise function:
[tex] \displaystyle \large{f(x) = \begin{cases} {x}^{3} \: \: (x < 0) \\ \sqrt[3]{x} \: \: (x > 0) \end{cases}}[/tex]
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:
[tex] \displaystyle \large{f(x) = \begin{cases} {x}^{3} \: \: (x < 0) \\ \sqrt[3]{x} \: \: (x > 0) \end{cases}}[/tex]
Since -8 is less than 0, we use x^3.
[tex] \displaystyle \large{f( - 8) = {( - 8)}^{3} } \\ \displaystyle \large{f( - 8) = - 512} [/tex]
Evaluate x = 8
From the piecewise function, since 8 is greater than 0, we use the cube root of x.
[tex] \displaystyle \large{f(8) = \sqrt[3]{8} }[/tex]
To evaluate the cube root, first we prime-factor the number.
[tex] \displaystyle \large{f(8) = \sqrt[3]{2 \cdot 2 \cdot 2} }[/tex]
Since it's a cube root, we pull three 2's out of the cube root and write only one 2.
[tex] \displaystyle \large{f(8) =2}[/tex]
Answer
Let me know if you have any questions!
