Question

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

Answer 1

So we have

`y = x^2ln(x/e^(x))`

Using log properties we can say it's

`y = x^2(ln(x) - ln(e^(x)))`
`y = x^2(ln(x) - x)`
`y = x^2ln(x) - x^3`

Differentiating we have

`dy/dx = x^2d/dxln(x) + ln(x)d/dxx^2 - 3x^2`

`dy/dx = x^2*1/x + ln(x)*2x - 3x^2`

`dy/dx = x + ln(x)*2x - 3x^2`

`dy/dx = x(1 + 2ln(x)) - 3x^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(1+2ln(x)) - 3x^2`

`3x^2 = x(1 + 2ln(x))`

Since we know `x != 0`, we can safely divide by x and see that

`3x = 1 + 2ln(x)`

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

`x > ln(x)`

And so

`3x > 3ln(x)`

And

`3ln(x) > 2ln(x) + 1`

So, there are no critical points.