Question

since there are no real roots for $y^{2}-2y + 4$, $g(y)$ exists for all real numbers. this means critical numbers occur where $0 = g(y)=\frac{y(4 - y)}{(y^{2}-2y + 4)^{2}}$. this happens only when the numerator equals 0. thus, the critical numbers are as follows. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)
$y=$

Explanation:

Step1: Set numerator equal to 0 (Critical numbers occur when numerator is 0)

$$y(4 - y)=0$$

Step2: Solve the equation

Using the zero - product property: if \(ab = 0\), then \(a = 0\) or \(b=0\).
For \(y(4 - y)=0\), we have two cases:

• Case 1: \(y=0\)
• Case 2: \(4 - y=0\), which gives \(y = 4\)

Answer:

\(0,4\)