How do you sketch the graph #y=ln(1/x)# using the first and second derivatives?

1 Answer
Jan 26, 2017

#f(x) = ln(1/x)# is monotone and strictly decreasing in its domain and therefore has no local extrema, and it is concave up everywhere.

Explanation:

Using the properties of logarithms we should see that:

#ln(1/x) = -lnx#

If we want to go through the whole sketcing process as an an exercise, we start by noting that #y=ln(1/x)# is defined and continuous for #x in (0,+oo)# and we analyze the limits at the boundaries of the domain:

#lim_(x->0^+) ln(1/x) = lim_(y->+oo) lny = +oo#

#lim_(x->+oo) ln(1/x) = lim_(y->0^+) lny = -oo#

Then we calculate the first and second derivatives:

#d/(dx) ln(1/x) = 1/(1/x) (-1/x^2) =-1/x#

#d^2/(dx^2) ln(1/x) = 1/x^2#

We can see that #f(x)# is monotone and strictly decreasing in its domain and therefore has no local extrema, and that it is concave up everywhere.

graph{ln(1/x) [-10, 10, -5, 5]}