For what values of x is f(x)=2x^4+4x^3+2x^2-2 concave or convex?

See solution below

Explanation:

Given function:

$f \left(x\right) = 2 {x}^{4} + 4 {x}^{3} + 2 {x}^{2} - 2$

$f ' \left(x\right) = 8 {x}^{3} + 12 {x}^{2} + 4 x$

$f ' ' \left(x\right) = 24 {x}^{2} + 24 x + 4$

The function will be concave when $f ' ' \left(x\right) \setminus \le 0$

$\setminus \therefore 24 {x}^{2} + 24 x + 4 \setminus \le 0$

$6 {x}^{2} + 6 x + 1 \setminus \le 0$

$\left(x + \setminus \frac{3 + \setminus \sqrt{3}}{6}\right) \left(x + \setminus \frac{3 - \setminus \sqrt{3}}{6}\right) \setminus \le 0$

$x \setminus \in \left(\setminus \frac{- 3 - \setminus \sqrt{3}}{6} , \setminus \frac{- 3 + \setminus \sqrt{3}}{6}\right)$

Hence, the given function is concave in $\left(\setminus \frac{- 3 - \setminus \sqrt{3}}{6} , \setminus \frac{- 3 + \setminus \sqrt{3}}{6}\right)$

the given function is convex in $\left(- \setminus \infty , \setminus \frac{- 3 - \setminus \sqrt{3}}{6}\right) \setminus \cup \left(\setminus \frac{- 3 + \setminus \sqrt{3}}{6} , \setminus \infty\right)$