Answer
The absolute maximum is $f(e^{1/2}) = \frac{1}{2e}$
The absolute minimum is $f(\frac{1}{2}) = -4~ln(2)$
Work Step by Step
$f(x) = x^{-2}~ln~x$
We can find the points when $f'(x) = 0$:
$f(x) = x^{-2}~ln~x$
$f'(x) = -2x^{-3}~ln~x+x^{-3} = 0$
$ln~x = \frac{1}{2}$
$x = e^{1/2}$
We can find the value of $f(e^{1/2})$ and the values of the function at the endpoints of the interval $[\frac{1}{2}, 4]$:
$f(e^{1/2}) = (e^{1/2})^{-2}~ln~(e^{1/2}) = \frac{1}{2e}$
$f(\frac{1}{2}) = (\frac{1}{2})^{-2}~ln~(1/2) = -4~ln(2)$
$f(4) = (4)^{-2}~ln~(4) = \frac{1}{16}~ln~4 \lt \frac{1}{2e}$
The absolute maximum is $f(e^{1/2}) = \frac{1}{2e}$
The absolute minimum is $f(\frac{1}{2}) = -4~ln(2)$
