Question

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)`?

Answer 1

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]}