Question

step 1
for $g(y)=\frac{y - 4}{y^{2}-2y + 8}$, we have
$g(y)=\frac{-y^{2}+8y}{(y^{2}-2y + 8)^{2}}\frac{-(y - 8)y}{(y^{2}-2y + 8)^{2}}$

step 2
critical numbers occur where $g(y)$ equals 0 or is undefined. $g(y)$ is undefined where the quadratic $y^{2}-2y + 8$ in the denominator is 0. so, $g(y)$ is undefined for the following values. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)
$y = dne$

step 3
since there are no real roots for $y^{2}-2y + 8$, $g(y)$ exists for all real numbers. this means critical numbers occur where $0 = g(y)=\frac{y(8 - y)}{(y^{2}-2y + 8)^{2}}$. 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: Recall critical - number condition

Critical numbers of a function occur where its derivative is 0 or undefined. We have $g'(y)=\frac{y(8 - y)}{(y^{2}-2y + 8)^{2}}$, and since the denominator $y^{2}-2y + 8$ has no real roots (using the discriminant $\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times8=4 - 32=-28<0$), $g'(y)$ is defined for all real $y$.

Step2: Set numerator equal to 0

We set the numerator of $g'(y)$ equal to 0. So, $y(8 - y)=0$.

Step3: Solve the equation

Using the zero - product property, if $y(8 - y)=0$, then either $y = 0$ or $8 - y=0$ (i.e., $y = 8$).

Answer:

$0,8$