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).
In order for the ball to be used in the game, it must be able to meet the minimum and maximum weight requirements. These are the limits of the weight of the ball. If it exceeds the maximum limit of 16 ounces or below the minimum limit of 14 ounces, the ball will not be approved.
So, by adding 1.5 ounces, that would mean that the initial weight of the ball did not reach the minimum limit. The initial weight of the ball, denoted as x, have two possible values. The first value is the initial weight plus added with 1.5 ounces would reach 14 ounces.
x + 1.5 = 14
x = 12.5 ounces
The other scenario is when the initial weight is added to reach the maximum requirement of 16 ounces
x + 1.5 = 16
x = 14.5 ounces
From both answer, we could conclude that the initial weight has to be 12.5 ounces. If the initial weight were 14.5 ounces to begin with, there should be no need for air. It cans till be approved to be used.
So,
We can notice that the graph of g, is translated 2 units to the left and 4 units up. We can express these changes with the following equation:
Answer:
( 8,11)
Step-by-step explanation:
When x = 8 the output is 7
The new function
f(x) +4
when x = 8
The output is f(8) +4= 7+4 = 11
( 8,11)
(x+3y) (x-3y) = x^2 9y^2
(x-y) (x+3y) = x^2+2xy-3y^2
(xy+9y+2) (xy-3) = x^2 y^2 +9xy^2-xy-27y-6
(x^2+3xy-2) ( xy+3) = x^3 y+3x^2 y^2 +7xy-6
(2xy+x+y) (3xy^2-y) = 6x^2 y^3 +3x^2 y^2 +xy^2-2xy^2-xy-y^2
(xy+3x+2) ( xy-9) = x^2 y^2 +3x^2 y -7xy-27x-18
