Originally Posted by

**el123** $\displaystyle f(x)= 2x^2 + \frac {8}{x^2} $

$\displaystyle f'(x)= 4x - \frac{16}{x^3} $

$\displaystyle f''(x) = 4+\frac{48}{x^4} $

So taking f''(x)

$\displaystyle \frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}$

$\displaystyle \frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4$

$\displaystyle 4x^4+48 = 0$

$\displaystyle \textcolor{red}{4x^4 + 48 = 0}$ has no real solutions.

f''(x) > 0 for all x in the domain of f, so f is concave up everywhere in its domain, i.e. no inflection points.