Answer
$c=0$ and $c=2$ are critical numbers of $g$.
Work Step by Step
How to find the critical numbers of a function according to definition
1) Find all numbers of $c$ satisfying whether $f'(c)=0$ or $f'(c)$ does not exist.
2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not.
- If $c$ lies in $D_f$, $c$ is a critical number of $f$.
- If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$.
$$g(y)=\frac{y-1}{y^2-y+1}$$
We see that $y^2-y+1\neq0$ for all values of $y$, so the denominator here cannot be $0$. Therefore, $$D_g=R$$
1) Now, we find $g'(y)$
$$g'(y)=\frac{1(y^2-y+1)-(2y-1)(y-1)}{(y^2-y+1)^2}$$
$$g'(y)=\frac{y^2-y+1-2y^2+2y+y-1}{(y^2-y+1)^2}$$
$$g'(y)=\frac{-y^2+2y}{(y^2-y+1)^2}$$
- First, $$g'(c)=0$$
$$-c^2+2c=0$$
$$-c(c-2)=0$$
$$c=0\hspace{0.5cm}or\hspace{0.5cm}c=2$$
There are no values of $c$ that makes $g'(c)$ not exist, for $(c^2-c+1)^2$ cannot equal $0$.
2) Examine whether $c$ lies in $D_g$ or not.
We see that both $c=0$ and $c=2\in R$, so both lie in $D_g$.
We conclude that $c=0$ and $c=2$ are critical numbers of $g$.
