What are the critical points of f(x) =x^2ln(x/e^x)?

Nov 17, 2015

So we have

$y = {x}^{2} \ln \left(\frac{x}{e} ^ \left(x\right)\right)$

Using log properties we can say it's

$y = {x}^{2} \left(\ln \left(x\right) - \ln \left({e}^{x}\right)\right)$
$y = {x}^{2} \left(\ln \left(x\right) - x\right)$
$y = {x}^{2} \ln \left(x\right) - {x}^{3}$

Differentiating we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \frac{d}{\mathrm{dx}} \ln \left(x\right) + \ln \left(x\right) \frac{d}{\mathrm{dx}} {x}^{2} - 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \cdot \frac{1}{x} + \ln \left(x\right) \cdot 2 x - 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x + \ln \left(x\right) \cdot 2 x - 3 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left(1 + 2 \ln \left(x\right)\right) - 3 {x}^{2}$

A critical point is a point in which the derivative is null, knowing that and that the domain of the derivative is the same as that of the function, we have

$0 = x \left(1 + 2 \ln \left(x\right)\right) - 3 {x}^{2}$

$3 {x}^{2} = x \left(1 + 2 \ln \left(x\right)\right)$

Since we know $x \ne 0$, we can safely divide by x and see that

$3 x = 1 + 2 \ln \left(x\right)$

However, we have that, for every value of $x$ greater than 0,

$x > \ln \left(x\right)$

And so

$3 x > 3 \ln \left(x\right)$

And

$3 \ln \left(x\right) > 2 \ln \left(x\right) + 1$

So, there are no critical points.