In the function ln(x)/sqrt(x) what does the root of the derivative mean? It's not an extrema, so what is it?

Question

anonymous · Answer

If its not an extrema, than it is a saddle point.

deoxna · Answer

Actually, upon consulting wolframalpha, it turns out it is a maximum: https://www.wolframalpha.com/input/?i=+ln%28x%29%2Fsqrt%28x%29 But why? That would mean that the function would begin decreasing after that and would eventually have a second root, but t apparently only has one root....

cruffo · Answer

One way is to check the sign of the slope on either side of the critical value (otherwise known as cv or root).

Positive slope on the left and negative slope on the right means that the cv is a maximum.

Let $f(x) = \frac{\ln(x)}{\sqrt{x}}$, then $f'(x) = \frac{2-\ln(x)}{2x^{3/2}}$.  Since $f(x)$ is not defined at $x=0$, we don't need to worry about the denominator of $f'(x)$.  Setting the numerator of $f'(x)$ equal to zero, we get a solution of $e^2 \approx 7$.

To check the slope of $f(x)$ we evaluate the first derivative:
$f'(5)$ is positive, meaning that $f(x)$ is increasing on the left of the cv.
$f'(8)$  is negative, meaning that $f(x)$ is decreasing on the right|dw:1356839861063:dw| of the cv.

So that means that the cv is a maximum :)

cruffo · Answer

Since there are no other critical values, this is the only extrema for $f(x)$.

deoxna · Answer

Yes, but that isn’t the whole story because f that were the only change n slope, then the function would eventually cross the x-axis and hence have a second root that is not accounted for. However, the function has an inflection point at e^(8/3) so that might explain why a second root doesn't exist.