Answer:
You're correct the answer is B) [tex]-5 \leq y \leq -1[/tex]
Step-by-step explanation:
The key behind this problem is think about the function [tex]\sqrt[3]{x}[/tex] and think what happen when you plug in values in it.
So we're going to try with some simply values for evaluate
[tex]\sqrt[3]{-8} = -2[/tex]
[tex]\sqrt[3]-1} = -1[/tex]
[tex]\sqrt[3]{0} = 0[/tex]
[tex]\sqrt[3]{1} = 1[/tex]
[tex]\sqrt[3]{8} = 2[/tex]
How you can see when plug in a negative number the function return a negative number, but when you plug in a positive number the function return a positive number (this isn't a proof of this happen in each value of the given interval, but is a good way of demonstrate the relation). This means that when you plug in a value this value is greater than the value of before, so the function is increasing it's outputs.
Now this is good because that's mean that the limits of the range are the limits of the domain evaluates in the function (because the smaller input give us the smaller output and the same with the greater). But the given function have different operation in it, so you have to interpret this operations of this way:
- When you have a negative value in the [tex]x[/tex] like in this occasion ([tex]\sqrt[3]{-x}[/tex]) the function reflex it's values with respect to the y axis (image put the functions in a mirror a draw the given image in the same cartesian plane).
- And you have a -3 this is move each value of the function 3 units down.
So with this information you evaluate each limit of the domain [tex]\{-8, 8\}[/tex] in the function and get the limits for the range.
[tex]f(-8) = \sqrt[3]{-(-8)} -3= \sqrt[3]{8} -3 = 2 -3 = -1[/tex]
[tex]f(8) = \sqrt[3]{-8} -3= -2 -3 = -5[/tex]
So the range of the function in the given interval is equal to [tex]\{-1, -5\}[/tex] in the interval notation is equal to [tex]-5 \leq y \leq -1[/tex]
