The preliminaries:

f'(x)=2x-a/x²

f"(x)=2+2a/x³

a.

A local extremum of a function continuous in an interval can occur at critical points OR at extremes of a closed interval. A critical point is defined as where f'(x)=0, or where there is discontinuity in f'(x)

In the given function, the domain of f(x) is (-∞,0)∪(0,∞) with the discontinuity at x=0 excluded.

A local minimum occurs at x=2 therefore when f'(2)=0 AND f"(2)>0.

f'(2)=0 =>

2x-a/x²=0

a=2(2)³

=16

Check if x=2 is a minimum:

f"(2)=2+2*16/2³

=2+4

=6 >0 therefore x=2 is a minimum.

b.

A *necessary* condition of a point of inflection occurs at x=c is when f'(c)=0, and f"(c)=0.

For a point of inflection to be located at x=1, solve for f"(1)=0,

2+2a/x³=0

2+2a/(1)³=0

a=-1

or

f(x)=x²-1/x

Check if point is an inflection point:

f"(1)=2+2a=0 OK

f"(0.9)=-0.74 <0 (concave down)

f"(1.1)=+0.49 >0 (concave up)

Thus x=1 is an inflection point.

Alternatively, calculate

f^{III}(x)=-6a/x^{4}

=6/x^{4}

f^{III}(1)=6 >0

therefore f(x) changes from concave down to concave up at x=1, therefore f(x)=x²-1/x has an inflection point at x=1.