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

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Answer 1 · 2017-01-24T20:54:03

For #x=1# the function has a local minimum.

Calculate the first derivative of the function:

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

For #x=1# we have:

#f'(x) = 0#

so in this point the function is stationary and neither increasing nor decreasing.

If we calculate the second derivative:

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

we can see that #f''(1) = 2 > 0# so in #x=1# the function has a local minimum.