Let ,R be such that the function f(x)=cases2(x^2-2)+2 x, & x

Question

Let $\alpha,\beta\in\mathbb{R}$ be such that the function $f(x)=\begin{cases}2\alpha(x^2-2)+2\beta x, & x<1\ (\alpha+3)x+(\alpha-\beta), & x\geq1\end{cases}$ is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to:

Accepted Answer

Continuity at $x=1$:
$$2\alpha(1-2)+2\beta(1)=(\alpha+3)+(\alpha-\beta) \implies -2\alpha+2\beta=2\alpha-\beta+3 \implies 3\beta-4\alpha=3 \quad\text{...(1)}$$

Differentiability at $x=1$: $f'(x)=4\alpha x+2\beta$ for $x<1$, $f'(x)=\alpha+3$ for $x>1$:
$$4\alpha+2\beta=\alpha+3 \implies 3\alpha+2\beta=3 \quad\text{...(2)}$$

Solving: multiply (1) by 2 and (2) by 3, then subtract: $17\alpha=3\implies \alpha=\dfrac{3}{17}$.

Substituting into (2): $2\beta=3-\dfrac{9}{17}=\dfrac{42}{17}\implies\beta=\dfrac{21}{17}$.

$$34(\alpha+\beta)=34\left(\frac{3}{17}+\frac{21}{17}\right)=34\cdot\frac{24}{17}=48$$

Solutions · 1