Critical number $y = \frac{1}{x^2 + 2}$ Seems pretty straight forward but my book seems to be giving an incorrect answer without any explanation to their magic.
$$y = \frac{1}{x^2 + 2}$$
I know that this has no 0 so that rule of finding a critical number can be discarded. 
$$y = (x^2 + 2)^{-1}$$
$$y' = -1(x^2 + 2)^{-2}\cdot 2x$$
$$y' = \frac{-2x}{(x^2 + 2)^{2}}$$
So now I am not so sure how my book got c = 0.
 A: $$y' = \frac{-2x}{(x^2+2)^2}$$
Critical points occur if and when there are values of $x$ which make  derivative of the function equal to zero. (I.e., you find the "zeros" of the function, if any.)
You also check for points at which the function is not differentiable. The denominator in both the function $y$ and the derivative $y'$ are defined (and positive) for all real numbers: The denominator $(x^2 + 2)^2 \gt 0$ for all $x$. 
So here, the only points we need to consider are the zeros of the derivative.
In this case, $\;y' = 0 \iff $ the numerator is zero. 
The numerator is zero if and only if $$-2x = 0 \iff x = 0$$ 
So there is indeed a critical point when $x = 0$, with the point being $(x,y) =  (0, 1/2)$. 
A: In your second line you have that the derivative is
$$\frac{-2x}{(x^2+2)^{2}}$$
which is 0 when x=0 so it is a critical point of this function. In the third line you omitted the x in the derivative.
A: A fraction is zero only if the numerator is zero and the denominator is not.  The denominator, $(x^2+2)^2$, is never zero no matter what number $x$ is.  So the question is: what must $x$ be in order that the numerator, $-2x$, be zero?  In other words, how to solve this equation for $x$:
$$
-2x=0.
$$
Divide both sides by $-2$:
$$
x = \frac{0}{-2}.
$$
Now the problem is: what do you get when you divide $0$ by $-2$?
