Question

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

Answer 1

`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]}