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

1 Answer
Jan 7, 2016

#f(x)# is neither increasing nor decreasing when #x=1#.

Explanation:

Find #f(1)#.

If #f(1)>0#, the function is increasing when #x=1#.
If #f(1)<0#, the function is increasing when #x=1#.
If #f(1)=0#, there is a critical value and the function is neither increasing nor decreasing.

To find #f(x)#, use the product rule, which states that for a function #f(x)=g(x)h(x)#, then

#f'(x)=g'(x)h(x)+h'(x)g(x)#

First, note that I'm assuming you meant #(ln(x))^2#, and not #ln(x^2)#.

#f'(x)=(ln(x))^2d/dx(x)+xd/dx((ln(x))^2)#

Find each derivative:

#d/dx(x)=1#

The next requires chain rule:

#d/dx((ln(x))^2)=2ln(x)d/dx(ln(x))=2ln(x)(1/x)#

Plug these back in to find #f'(x)#.

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

Simplify.

#f'(x)=(ln(x))^2+2ln(x)#

Now, find #f'(1)#.

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

Thus, #f(x)# is neither increasing nor decreasing when #x=1#.

This is the graph of #f(x)#:

graph{x(ln(x))^2 [-3.16, 10.887, -1.46, 5.565]}

There is a minimum when #x=1#.