Let α, β ∈ R be such that the function f(x)= {2 α(x^2-2)+2 β x, x<1; (α+3) x+(α-β), x ≥ 1} be differentiable at all x ∈ R. Then 34(α+β) is equal to

Let $\displaystyle \alpha, \beta \in \mathbb{R}$ be such that the function $\displaystyle f(x)= \begin{cases}2 \alpha\left(x^2-2\right)+2 \beta x & , x<1 \\ (\alpha+3) x+(\alpha-\beta) & , x \geq 1\end{cases}$ be differentiable at all $\displaystyle x \in \mathbb{R}$. Then $\displaystyle 34(\alpha+\beta)$ is equal to