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$Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) {:[g(y)=(y-1)/(y^(2)-3y+3)],[y=◻]:}$

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) $\newline$ $\begin{array}{l} g(y)=\frac{y-1}{y^{2}-3 y+3} \\ y=\square \end{array}$

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Math Problems

Algebra 2

Complex conjugate theorem

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Q. Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

\newline

\begin{array}{l} g(y)=\frac{y-1}{y^{2}-3 y+3} \\ y=\square \end{array}

Find Critical Numbers: To find the critical numbers, we need to find where the derivative of $g(y)$ is zero or undefined. $\newline$ Take the derivative of $g(y)$ using the quotient rule: $\frac{v(u') - u(v')}{v^2}$ , where $u = y - 1$ and $v = y^2 - 3y + 3$ .
Calculate Derivative: Calculate the derivative of $u$ : $u' = (y - 1)' = 1$ .
Apply Quotient Rule: Calculate the derivative of $v$ : $v' = (y^2 - 3y + 3)' = 2y - 3$ .
Simplify Numerator: Apply the quotient rule: $g'(y) = \frac{(y^2 - 3y + 3)(1) - (y - 1)(2y - 3)}{(y^2 - 3y + 3)^2}$ .
Set Equal to Zero: Simplify the numerator: $g'(y) = \frac{y^2 - 3y + 3 - 2y^2 + 3y - 3}{(y^2 - 3y + 3)^2}$ .
Solve for $y$ : Combine like terms in the numerator: $g'(y) = \frac{-y^2 + 3}{(y^2 - 3y + 3)^2}$ .
Take Square Root: Set the numerator equal to zero to find where $g'(y)$ is zero: $-y^2 + 3 = 0$ .
Take Square Root: Set the numerator equal to zero to find where $g'(y)$ is zero: $-y^2 + 3 = 0$ . Solve for $y$ : $y^2 = 3$ .
Take Square Root: Set the numerator equal to zero to find where $g'(y)$ is zero: $-y^2 + 3 = 0$ . Solve for $y$ : $y^2 = 3$ . Take the square root of both sides: $y = \pm\sqrt{3}$ .