Answer:
5
Step-by-step explanation:
The given rational function is
[tex] \frac{2 {y}^{2} + 2}{ {y}^{3} - 5 {y}^{2} + y - 5} [/tex]
This function is not defined when the denominator is zero.
To find the restrictions, we equate the denominator to zero and solve for y.
[tex] {y}^{3} - 5 {y}^{2} + y - 5 = 0[/tex]
We factor to get;
[tex] {y}^{2} (y - 5) + 1(y - 5) = 0[/tex]
Factor further:
[tex]( {y}^{2} + 1) (y - 5) = 0[/tex]
[tex] {y}^{2} = - 1 \: or \: y = 5[/tex]
The real solution is
[tex]y = 5[/tex]
The restriction is
[tex]y = 5[/tex]
