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

Jan 12, 2017

$f \ge 2$. Symmetrical about y-axis.
Minimum : 2. No maximum. No POI ( point of inflexion ).
Convex towards origin.

#### Explanation:

$f \to \pm \infty$, as $x \to 0 \mathmr{and} \pm \infty$.

$f > 0$, with the minimum given by

$f ' = 2 x - \frac{2}{x} ^ 3 = 0$, giving zeros x=+-1#.

The minimum f is $f \left(\pm 1\right) = 2$, twice.

$f ' ' = 2 + \frac{6}{x} ^ 4 > 0$.

$x = 0 \uparrow$ is the asymptote.

On the left (${Q}_{2}$) and the right of the asymptote, (${Q}_{1}$),

f' is increasing, from $- \infty \to \infty$, revealing convexity,

in respect of each branch of the graph, in ${Q}_{1} \mathmr{and} {Q}_{2}$..

So, f is minimum 2 at $x = \pm 1$ and there is no point of inflexion.

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