Find critical numbers:

#f'(x) = 2-3/x^2 = (2x^2-3)/x^2#

#f'(x) = 0# at #x = +-sqrt(3/2)# and #f'(x)# DNE at #x=0#.

#0# is not in the domain of #f#, so the only critical numbers ar #+-sqrt(3/2)#

**First Derivative Test for Local Extrema**

On #(-oo, -sqrt(3/2))#, #f'# is positive

On #(-sqrt(3/2), 0)#, #f'# is negative #" "# so #f(-sqrt(3/2))# is a relative maximum.

On #(0, sqrt(3/2))#, #f'# is negative

On #(sqrt(3/2), oo)#, #f'# is positive #" "# so #f(sqrt(3/2))# is a relative minimum.

**Second Derivative Test for Local Extrema**

#f''(x) = 6/x^3#

#f''(-sqrt(3/2))# is negative, so #f(-sqrt(3/2))# is a relative maximum.

#f''(-sqrt(3/2))# is positive, so #f(-sqrt(3/2))# is a relative minimum.

**Finish by doing arithmetic**

Find the values by doing the arithmetic #f(-sqrt(3/2))# and #f(sqrt(3/2))#. (Actually, we only need to do one of these, because #f# is an odd function.)