find the critical numbers of the function.\ng(y)=\\frac{y - 2}{y^{2}-2y + 4}\nstep 1\nfor g(y)=\\frac{y…

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if (g(y)=\frac{u(y)}{v(y)}), then (g^{\prime}(y)=\frac{u^{\prime}(y)v(y)-u(y)v^{\prime}(y)}{v(y)^{2}}). Here, (u(y)=y - 2), so (u^{\prime}(y)=1); (v(y)=y^{2}-2y + 4), so (v^{\prime}(y)=2y-2). [ \begin{align*} g^{\prime}(y)&=\frac{1\times(y^{2}-2y + 4)-(y - 2)\times(2y-2)}{(y^{2}-2y + 4)^{2}}\ \end{align*} ]

Step2: Expand the numerator

Expand ((y - 2)\times(2y-2)=2y^{2}-2y-4y + 4=2y^{2}-6y + 4). Then (u^{\prime}(y)v(y)-u(y)v^{\prime}(y)=(y^{2}-2y + 4)-(2y^{2}-6y + 4)). [ \begin{align*} (y^{2}-2y + 4)-(2y^{2}-6y + 4)&=y^{2}-2y + 4-2y^{2}+6y - 4\ &=-y^{2}+4y \end{align*} ]

Step3: Write the final derivative

So (g^{\prime}(y)=\frac{-y^{2}+4y}{(y^{2}-2y + 4)^{2}}=\frac{-y(y - 4)}{(y^{2}-2y + 4)^{2}})

Answer:

(g^{\prime}(y)=\frac{-y(y - 4)}{(y^{2}-2y + 4)^{2}})