Given f(x) = x²ln (ax) with a ≠0, find the set of extreme

Given $f(x) = x^{2}\ln\bigl(ax\bigr)$ with $a\neq0$, find the set of extreme points.

Step 1: Domain

The logarithm requires $ax\gt 0$, so the domain is $\bigl(0,+\infty\bigr)$ when $a\gt 0$ and $\bigl(-\infty,0\bigr)$ when $a\lt 0$. In either case $x\neq0$.

Step 2: The derivative

\[ f'(x) = 2x\ln\bigl(ax\bigr)+x^{2}\cdot\frac{a}{ax} = 2x\ln\bigl(ax\bigr)+x = x\left(2\ln\bigl(ax\bigr)+1\right). \tag{1} \]

Step 3: The critical point

On the domain $x\neq0$, so the second factor must vanish: \[\ln\bigl(ax\bigr) = -\frac12 \;\Longleftrightarrow\; ax = e^{-1/2} \;\Longleftrightarrow\; x_{0} = \frac{1}{a\sqrt{e}}, \tag{2}\] which satisfies $ax_{0} = e^{-1/2}\gt 0$ and so always lies in the domain, whatever the sign of $a$.

Step 4: It is a minimum

\[ f''(x) = \left(2\ln\bigl(ax\bigr)+1\right)+x\cdot\frac2x = 2\ln\bigl(ax\bigr)+3, \qquad f''\bigl(x_{0}\bigr) = -1+3 = 2\gt 0. \]

The sign of $f'$ confirms it directly: the factor $x$ keeps a fixed sign on the domain, while $2\ln(ax)+1$ increases through zero at $x_{0}$ when $a\gt 0$ and decreases through zero when $a\lt 0$ — in both cases $f'$ passes from $-$ to $+$.

Step 5: The minimum value

\[ f\bigl(x_{0}\bigr) = x_{0}^{2}\ln\bigl(ax_{0}\bigr) = \frac{1}{a^{2}e}\left(-\frac12\right) = -\frac{1}{2a^{2}e}. \]

Step 6: Verification

\[a = 1:\ x_{0} = 0.6065307,\ f\bigl(x_{0}\bigr) = -0.1839397 = -\frac{1}{2e} \;\;\checkmark\]

\[a = -2:\ x_{0} = -0.3032653,\ f\bigl(x_{0}\bigr) = -0.0459849 = -\frac{1}{8e} \;\;\checkmark\]

\[\text{Eliminating } a:\ a^{2} = \frac{1}{e\,x_{0}^{2}} \;\Longrightarrow\; f\bigl(x_{0}\bigr) = -\frac{x_{0}^{2}}{2},\ \text{so all minima lie on } y = -\frac{x^{2}}{2} \;\;\checkmark\]