Properties of the function

Problem

The function \(f:x\mapsto\frac{\ln x}{x^2}\) with maximal domain \(\mathbb{D}\) is given.

1. Give \(\mathbb{D}\) as well as the roots of \(f\), and determine \(\lim\limits_{x\rightarrow0}f(x)\).

2. Determine the \(x\)-coordinate of the point in which the graph of \(f\) has a horizonal tangent line.

Solution of part a

The logarithm is defined for arguments \(x>0\) only. The denominator \(x^2\) contributes a gap in the domain at \(x=0\). The maximal domain is hence given by

\[\mathbb{D}=]0;\infty[\,.\]

We obtain the roots from the roots of the numerator:

\[\ln(x) = 0\Rightarrow x=1\,.\]

We plot the function with Sage.

We can also verify the root with Sage.

As the graph produced by Sage suggests, the function goes to \(-\infty\) as \(x\rightarrow0\). This can also be established by the fact that on the one hand the enumerator goes to \(-\infty\) and on the other hand the denominator of the function goes to \(0^+\).

Solution of part b

A horizontal tangent line corresponds to an extremum of the function. To identify such a point, we have to determine the derivative first and, subsequently, set it equal to 0:

\[\begin{split}f'(x) = \frac{1-2\ln(x)}{x^3} \overset{!}{=} 0 \Rightarrow \ln x= \frac{1}{2}\\\end{split}\]

This yields a horizontal tangent line at

\[x = e^{\frac{1}{2}} = \sqrt{e}\]

which we add to the sketch of the function: