Question

Help me please???
I don’t know how to get started!

Answer 1

A differentiable function
`f(x)` is increasing on an open interval
`(a,b)` if
`f'(x)>0` for all
`a<x<b`, and decreasing if
`f'(x)<0`. For this problem, you then need to compute the derivative:

`f(x)=x^2\ln x\implies f'(x)=2x\ln x+x=(2\ln x+1)x`

then solve for
`f'(x)=0`:

`(2\ln x+1)x=0\implies x=0\text{ or }x=e^(-1/2)`

We can ignore
`x=0` because
`x^2\ln x` is defined only for
`x>0`. So we have two intervals to consider,
`(0,e^(-1/2))` and
`(e^(-1/2),\infty)`. All we need to do is pick any value from either interval and check the sign of the derivative
`f'(x)`. Since
`e^(-1/2)\approx0.606`, from the first interval we can take
`x=\frac12`, and from the second we can pick
`x=1`.

`f'\left(\frac12\right)\approx-0.193<0`

`f'(1)=1>0`

The above indicates that
`f(x)` is decreasing on the first interval
`(0,e^(-1/2))`, and increasing on the second interval
`(e^(-1/2),\infty)`.

For part (b), we use the info from above as part of the first derivative test for extrema. We have one critical point at
`x=e^(-1/2)`, and we know how
`f(x)` behaves to either side of this point;
`f(x)` decreases to left of it, and increases to the right. This pattern is indicative of a minimum occurring at
`x=e^(-1/2)`, and we find that
`f(x)` has the (local) minimum value of
`f(e^(-1/2))=-\frac1{2e}\approx-0.184`.