Is #f(x)=x^2lnx# increasing or decreasing at #x=1#?

1 Answer
Jan 17, 2016

Increasing.

Explanation:

The derivative of a function can be used to determine if a function is increasing or decreasing at a point.

  • If #f'(1)<0#, then #f(x)# is decreasing at #x=1#.
  • If #f'(1)>0#, then #f(x)# is increasing at #x=1#.

First, find #f'(x)# through the product rule.

#f'(x)=lnxd/dx[x^2]+x^2d/dx[lnx]#

#f'(x)=lnx*(2x)+x^2(1/x)#

#f'(x)=2xlnx+x#

Now, find #f'(1)#.

#f'(1)=2(1)ln(1)+1#

#f'(1)=2(0)+1#

#f'(1)=1#

Since #f'(1)>0#, the function is increasing at #x=1#.

We can check a graph of #f(x)#:

graph{x^2lnx [-0.221, 1.6745, -0.353, 0.595]}