# Is f(x)=sqrt(ln(x)^2) increasing or decreasing at x=1?

Mar 25, 2016

x = 1 meets the two branches of log graphs representng this equation, at the common point (1. 0). This is a node for the graph. f(x) is increasing for one branch and decreasing for the other.

#### Explanation:

Importantly, the statement $\sqrt{{a}^{2}} = \pm a$ is slashed as wrong by quite many. Here is another example for elucidation on this true statement.

$\sqrt{\ln} {\left(x\right)}^{2} = \pm \ln x$

.So, the given equation is the combined equation for the pair
y = f(x) = + ln x and y = $-$ln x, x > 0..

For x > 1, the graph for the first is above the x-axis It is below the x-axis, for 0 < x < 1,
For the graph of the second equation in the pair, it is vice versa.

y-axis ( x = 0 ) is the asymptote.

Separately, f'(x) = $= \pm \frac{1}{x}$. For every x > 0, this is indeed the derived relation for the combined (given) equation

Anyway, at x = 1, y = 0 and f'(1) = +1 > 0 for one branch, and $- 1$< 0 for the other. So, one way it is increasing and, in the other way, it is decreasing.