Question

find the critical numbers of the function.
g(y)=\frac{y - 2}{y^{2}-2y + 4}
step 1
for g(y)=\frac{y - 2}{y^{2}-2y + 4}, we have
g(y)=\square

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 \(v^{\prime}(y)=2y-2\).

$$
LATEXBLOCK0
$$

Step2: Expand the numerator

Expand \((y - 2)\times(2y-2)=2y^{2}-2y-4y + 4=2y^{2}-6y + 4\).
Then \(u^{\prime}(y)v(y)-u(y)v^{\prime}(y)=(y^{2}-2y + 4)-(2y^{2}-6y + 4)\).

$$
LATEXBLOCK1
$$

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:

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