Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.
To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.
Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.
Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.
Therefore:
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 8} \\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%208%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
The nth root of a can contain negative number only if n is an odd number.
![\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 2 \times - 2 \times - 2} \\ f(4) = 3 \end{cases}} \\ \large{ \begin{cases} f( - 8 ) = - 2\\ f(4) = 3 \end{cases}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20%20%5Csqrt%5B3%5D%7B%20-%202%20%5Ctimes%20-%20%202%20%5Ctimes%20%20%20-%202%7D%20%20%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D%20%5C%5C%20%20%5Clarge%7B%20%20%5Cbegin%7Bcases%7D%20f%28%20-%208%20%29%20%3D%20%20-%202%5C%5C%20f%284%29%20%3D%203%20%5Cend%7Bcases%7D%7D)
Answer
Answer:
unbiased, I think
Step-by-step explanation:
Positive six and negative sixteen