How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for F(x)= (x^2)(lnx)?

1 Answer
Jan 3, 2017

By analysing F'(x) over the domain of F(x): x>0
F(x) has an absolute minimum at x=1/sqrt(e)
F(x) is decreasing for 0< x<1/sqrt(e) and increasing for x>1/sqrt(e)

Explanation:

F(x) = x^2lnx

F'(x) = x^2.!/x + lnx*2x (Product Rule)

= x+2x*lnx

For local extrema F'(x) =0

Thus: x+2xlnx = 0

x(1+2lnx)=0

x=0 or lnx=-1/2

Since F(x) is defined for x>0 our local extrema will be at:

lnx=-1/2 -> x=e^(-1/2) = 1/sqrt(e) ~= 0.60653

Hence:
F(x) has an absolute minimum at x=1/sqrt(e)
F(x) is decreasing for x<1/sqrt(e) and increasing for x>1/sqrt(e)

This is shown by the graph of F(x) below:

graph{x^2*lnx [-0.542, 2.877, -0.449, 1.26]}