Find all points of contact of horizontal tangents to the curve $y = 2\sqrt x + \frac 1{ \sqrt x}$ Find all points of contact of horizontal tangents to the curve 
$$y = 2\sqrt x + \frac 1{ \sqrt x}$$  
I found the derivative:
$$\frac {dy}{dx} = x^{-1/2} - \frac 12x^{-3/2}$$
Which can be simplified down to
$$ \frac {dy}{dx} = \frac  1{\sqrt x} - \frac 1 {2x\sqrt x}$$
Then I used the Null Factor Law:
$\frac{1}{\sqrt{x}} = 0$ or $\frac{1}{2x \sqrt{x}} = 0$.
I got stuck here.
 A: The Null Factor Law does not apply here. The law only applies to products, which, in the case of one multiplicand being $0$, the whole expression is $0$. This is not the case with addition; the law does not apply here.
When finding horizontal tangent lines, you set the derivative equal to zero, then solve for $x$:
$$\dfrac{1}{\sqrt{x}} - \dfrac{1}{2x\sqrt x} = {2x - 1 \over 2x\sqrt x} \Rightarrow {2x - 1 \over 2x\sqrt x} = 0$$
Since we're setting the derivative equal to zero, we're looking at when the numerator of the fraction is $0$ (and thus the whole expression is $0$):
$$2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \dfrac{1}{2}$$
Therefore, the graph of $2\sqrt x - \frac 1{\sqrt x}$ has a horizontal tangent line at $x = \frac{1}{2}$.
A: You are not using the Null Factor Law properly.  You are taking two terms, not factors, and setting them to zero.  However, you can pull a factor of x^(-3/2) from your derivative:
$$\frac{dy}{dx}=x^\frac{-3}{2}(x-\frac{1}{2})$$
Equating both its factors to zero, we get $x=0$ or $x=1/2$.  The $x=0$ answer is fake as the initial expression will get a zero denominator.  Therefore, the one true horizontal tangent happens at $x=1/2$, whose y value turns out to be $2\sqrt2$.
Ans: $$(\frac{1}{2}, 2\sqrt{2})$$
