Dannng I thought I seen this one before
To get the extrema, derive the function.
You get y' = 2x^-1/3 - 2.
Set this equal to zero, and you get x=0 as the location of a critical point.
Since you are on a closed interval [-1, 1], those points can also have an extrema.
Your min is right, but the max isn't at (1,1). At x=-1, you get y=5 (y = 3(-1)^2/3 -2(-1); (-1)^2/3 = 1, not -1).
Thus, the maximum is at (-1, 5).
Answer:
32.both of their ideas are correct whether it is to replace x or y first it doesn't matter only one can be replaced to find the value of the other.
33.It doesn't really matter all of them are correct.
Answer:
Which points are collinear?
G, A, and T
G, E, and T
A, T, and E
G, A, and E
Step-by-step explanation:
that what i nkow sorry
