since there are no real roots for $y^{2}-2y + 4$, $g(y)$ exists for all real numbers. this means critical…

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.)\n$y=$

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.)\n$y=$

Answer

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)