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

Question

Samantha Adkison · Accepted Answer

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]}

Emma Aldridge · Answer

undefined To find the extrema and saddle points of $ f(x) = 2x^2 \ln(x) $, we first need to find its critical points by setting the derivative equal to zero and solving for $ x $. Then, we'll use the second derivative test to determine the nature of these critical points.

The first derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = 4x \ln(x) + \frac{2x}{x} = 4x \ln(x) + 2 $$

Setting $ f'(x) = 0 $, we get:

$$ 4x \ln(x) + 2 = 0 $$

$$ 4x \ln(x) = -2 $$

$$ \ln(x) = -\frac{1}{2} $$

$$ x = e^{-\frac{1}{2}} $$

Now, let's find the second derivative of $ f(x) $ to use the second derivative test:

$$ f''(x) = 4 \ln(x) + 4x \times \frac{1}{x} = 4 \ln(x) + 4 $$

At $ x = e^{-\frac{1}{2}} $, $ f''(x) $ is:

$$ f''(e^{-\frac{1}{2}}) = 4 \ln(e^{-\frac{1}{2}}) + 4 = 4 \times (-\frac{1}{2}) + 4 = -2 + 4 = 2 $$

Since $ f''(e^{-\frac{1}{2}}) > 0 $, the function has a local minimum at $ x = e^{-\frac{1}{2}} $. Hence, $ (e^{-\frac{1}{2}}, f(e^{-\frac{1}{2}})) $ is the point of minimum.

There are no other critical points, so there are no additional extrema or saddle points for this function.