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.
Best Answer
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 $x = \frac{1}{2}$.