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
#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