How do you find the maximum, minimum and inflection points and concavity for the function #f(x) = x^2 + 1/x^2#?

1 Answer
Jan 12, 2017

#f >=2#. Symmetrical about y-axis.
Minimum : 2. No maximum. No POI ( point of inflexion ).
Convex towards origin.

Explanation:

#f to +-oo#, as #x to 0 and +-oo#.

#f > 0#, with the minimum given by

#f'=2x-2/x^3=0#, giving zeros x=+-1#.

The minimum f is #f(+-1)=2#, twice.

#f''=2+6/x^4 > 0#.

#x = 0 uarr# is the asymptote.

On the left (#Q_2#) and the right of the asymptote, (#Q_1#),

f' is increasing, from #-oo to oo#, revealing convexity,

in respect of each branch of the graph, in #Q_1 and Q_2#..

So, f is minimum 2 at #x = +-1# and there is no point of inflexion.

graph{(y-2)(y-x^2-1/x^2)=0 [-20, 20, -10, 10]}