Answer:
0, 10
Step-by-step explanation:
The given function is:
[tex]g(y) = \frac{y-5}{y^2-3y+15}[/tex]
According to the quotient rule:
[tex]d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}[/tex]
Applying the quotient rule:
[tex]g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}[/tex]
The values for which g'(y) are zero are the critical points:
[tex]g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10[/tex]
The critical values are y = 0 and y = 10.
