What are the local extrema, if any, of $f (x) =(xlnx)^2/x$ ?

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Answer 1 · 2017-07-21T14:09:38

$f_min =f(1) =0$
$f_max =f(e^(-2)) approx 0.541$

$f(x) = (xlnx)^2/x$

$= (x^2*(lnx)^2)/x$

$= x(lnx)^2$

Applying the product rule

$f'(x) = x*2lnx*1/x + (lnx)^2*1$

$= (lnx)^2 + 2lnx$

For local maxima or minima: $f'(x) = 0$

Let $z= lnx$

$:. z^2 +2z = 0$

$z(z+2) =0 -> z=0 or z=-2$

Hence for local maximum or minimum:

$lnx = 0 or lnx = -2$

$:.x= 1 or x=e^-2 approx 0.135$

Now examine the graph of $x(lnx)^2$ below.

graph{x(lnx)^2 [-2.566, 5.23, -1.028, 2.87]}

We can observe that simplified $f(x)$ has a local minimum at $x=1$ and a local maximum at $x in (0, 0.25)$

Hence: $f_min =f(1) =0$ and $f_max =f(e^(-2)) approx 0.541$

What are the local extrema, if any, of f (x) =(xlnx)^2/xf (x) =(xlnx)^2/x?