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.
Answer:
m1 = 68°
m2 = 53°
m3 = 127
m4 = 31°
Step-by-step explanation:
m1 = 90-22
m2 = 68+59=127, 180-127
m3 = 180 - 53
m4 = 127+22=149, 180-149
Answer:
Your answer is
b.9
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Step-by-step explanation:
