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:

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

Let us study the sign of the first derivative:

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

$$f'(x)>0 \Longleftrightarrow x < 0 x > e^{-1/2}.$$

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.

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

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

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

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

Thus there is no oblique asymptote.

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

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

therefore:

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

Now let us compute the second derivative:

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

We have:

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

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

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