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

##### 1 Answer

#### Explanation:

Find

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

*First, note that I'm assuming you meant* *and not*

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

Simplify.

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

Now, find

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

Thus,

This is the graph of

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

There is a minimum when