Polynomial and natural logarithm.

$ f(x)= x^2 \ln x$ .

This function is defined on $ (0,+\infty)$ .

The function $ f$ is the product of the natural logarithm by a monomial, hence it is differentiable on its domain. We have:

$\displaystyle \forall x \in (0,+ \infty ), \; f'(x)= 2x \ln x + x^2 \cdot \frac 1x = 2x \ln x + x = x (2 \ln x +1 )$    

Let us study the sign of the first derivative:

$\displaystyle f'(x)=0$ $\displaystyle \Longleftrightarrow x=0$    or $\displaystyle \ln x = -\frac 12 \Longleftrightarrow x=0$    or $\displaystyle x = e^{-1/2}.$    
$\displaystyle f'(x)>0$ $\displaystyle \Longleftrightarrow x < 0$    or $\displaystyle x > e^{-1/2}.$    

Recall that $ f$ is defined over $ (0,+\infty)$ , thus the first derivative vanishes only at $ e^{-1/2}$ ; it is negative over $ (0, e^{-1/2} )$ and positive over $ (e^{-1/2}, +\infty )$ . The function $ f$ decreases over $ (0, e^{-1/2} )$ and increases over $ (e^{-1/2}, +\infty )$ . It has a minimum at $ e^{-1/2}$ .

Now let us check the limits of $ f$ at the open ends of its domain.

\begin{align*}\begin{cases}\underset{x \rightarrow + \infty}{\lim} x^2 = +\infty  \underset{x \rightarrow + \infty}{\lim} \ln x = +\infty \end{cases}\end{align*} $\displaystyle \Longrightarrow \underset{x \rightarrow + \infty}{\lim} f(x)= +\infty .$    
\begin{align*}\begin{cases}\underset{x \underset{>} \rightarrow 0}{\text{lim}} x...
...nderset{x \underset{>} \rightarrow 0}{\text{lim}} x =0^+ \end{cases}\end{align*} $\displaystyle \Longrightarrow \underset{x \underset{>} \rightarrow 0}{\text{lim}} f(x)=0^-.$    

Does the graph of the function $ f$ have an oblique asymptote:

$\displaystyle \frac {f(x)}{x}= x \ln x \Longrightarrow \underset{x \rightarrow + \infty}{\lim} \frac {f(x)}{x}= + \infty.$    

Thus there is no oblique asymptote.

Now, let us check how the graph looks like near the origin: We have

$\displaystyle f'(x)=2\underbrace{x \ln x}_{\rightarrow 0} + \underbrace{x}_{\rightarrow 0},$    

therefore:

$\displaystyle \underset{x \underset{>} \rightarrow 0}{\text{lim}} f'(x)= 0.$    

Now let us compute the second derivative:

$\displaystyle \forall x \in (0, + \infty ), \; f''(x)=2 \ln x + 3.$    

We have:

$\displaystyle f''(x) =0$ $\displaystyle \Longleftrightarrow \ln x = - \frac 32 \Longleftrightarrow x=e^{-3/2}.$    
$\displaystyle f''(x) > 0$ $\displaystyle \Longleftrightarrow \ln x > - \frac 32 \Longleftrightarrow x >e^{-3/2}.$    

Thus $ f$ is concave over $ (0, e^{-3/2} )$ and convex over $ ( e^{-3/2}, + \infty )$ . It has a point of inflexion at $ e^{-3/2}$ .

The graph of $ f$ is displayed in Figure 3.

Figure: The graph of $ f(x)= x^2 \ln x$ .
\begin{figure}\mbox{\epsfig{file=graph3.eps,height=6cm}}\end{figure}

Noah Dana-Picard 2007-12-28