Is $f (x) = x {(ln x)}^{2}$ $f(x)=x(lnx)^2$ increasing or decreasing at $x = 1$ $x=1$ ?

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Answer 1 · 2017-01-24T20:54:03

For $x = 1$ $x=1$ 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.

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