y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
N= night c= child
sandy= $10+$5c
kimmi= $25n
if sandy babysits 3 children for one night they will both be the same price
sandy= $10+5(3)=$25
kimmi= $25
Answer:
x+20 * 3
Step-by-step explanation:
* represents a multiplication symbol
Answer:
- 2x + 4 = 36
- 2x = 36-4
- 2x = 32
- x = 32/2
- x = 16
Step-by-step explanation:
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