Question

Can anybody explain how to solve this?

Answer 1

A. y^3<=16y -> y(y^2-16)<=0 -> y(y+4)(y-4)<=0 -> y<=-4 or 0<=y<=4
not the right answer.
B. 4y-y^2 >=0 -> y(4-y)>=0 -> 0<=y<=4 not the right answer
C y^3-16y>=0 -> y(y+4)(y-4)>=0 -> -4<=y<=0 or y>=4 RIGHT!
D y^2>=4y -> y^2-4y >=0 -> y(y-4)>=0 -> 0<=y<=4 not the right answer.

Answer 2

The Answer is C.

You know that by the graph: plug in the values between 0 and -4. they are all greater than or equal to 0.

If you try the values that aren't in between the inequality doesn't hold for example:

\[1^{3}-16(1) = 1-16 = -15 \ge 0\]

Which doesn't hold up. -15 isn't greater than or equal to 0, if you keep going until you reach 4 which finally gives you \[0 \le 0\] which you get the same thing. because of the y^3 term finally gets larger than the -16y term.

Answer 3

Thank you so much!!!