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

1 Answer
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.


Importantly, the statement #sqrt(a^2)=+-a# is slashed as wrong by quite many. Here is another example for elucidation on this true statement.

#sqrtln(x)^2 =+-lnx#

.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) = #=+-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.