1) [tex]f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = 2x^2 -4x + 5.\\f^{'}(x) = 4x -4\\f^{''}(x) = 4, f^{''}(x) > 0, \forall x \in \mathbb{R} \implies f \text{ e convexa pentru orice } x \in \mathbb{R}.[/tex]
2)
[tex]f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = 2x^3 -3x^2 + 5\\f^{'}(x) = 6x^{2} - 6x\\f^{''}(x) = 12x - 6, f^{''}(x) = 0 \implies x = \frac{1}{2}.\\f^{''}(0) = -6, f^{''}(1) = 6[/tex]
[tex]\text{Deci }f \text{ are punctul de inflexiune $x=\frac{1}{2}$ si e convexa pe intervalul $[\frac{1}{2}, \infty)$ si concava pe $(-\infty, \frac{1}{2}]$}[/tex]
