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
fIII(x)=-6a/x4
=6/x4
fIII(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.
