Question

What are the extrema and saddle points of `f(x)=2x^2 lnx`?

Answer 1

The domain of definition of:

`f(x) = 2x^2lnx`

is the interval `x in (0,+oo)`.

Evaluate the first and second derivatives of the function:

`(df)/dx = 4xlnx +2x^2/x = 2x(1+2lnx)`

`(d^2f)/dx^2 = 2(1+2lnx)+2x*2/x = 2+4lnx+4 = 6+lnx`

The critical points are the solutions of:

`f'(x) = 0`

`2x(1+2lnx) = 0`

and as `x > 0`:

`1+2lnx =0`

`lnx = -1/2`

`x =1/sqrt(e)`

In this point:

`f''(1/sqrte) = 6-1/2 = 11/2 > 0`

so the critical point is a local minimum.

The saddle points are the solutions of:

`f''(x) = 0`

`6+lnx =0`

`lnx = -6`

`x= 1/e^6`

and as `f''(x)` is monotone increasing we can conclude that `f(x)` is concave down for `x < 1/e^6` and concave up for `x > 1/e^6`

graph{2x^2lnx [-0.2943, 0.9557, -0.4625, 0.1625]}