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Find all the critical values

Find all the critical values-example-1

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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.
User Ramdaz
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