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

Question

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

Accepted Answer

$f(x)= $$2\alpha ({x}^{2}-2)+2\beta x, \& xGiven $f(x)$ is differentiable at all $x\in R$.$$

This means $f(x)$ is continuous and differentiable at $x=1$.

Continuity at $x=1$

$$$\begin{gathered} ==f(1) \ \underset{x\to {1}^{-}}{lim}f(x)=\underset{x\to {1}^{+}}{lim}f(x) \ \underset{x\to {1}^{-}}{lim}[2\alpha ({x}^{2}-2)+2\beta x]=\underset{x\to {1}^{+}}{lim}[(\alpha +3)x+(\alpha -\beta )] \ 2\alpha (1-2)+2\beta (1)=\alpha +3+\alpha -\beta \ -2\alpha +2\beta =2\alpha -\beta +3 \ 3\beta -4\alpha =3-(1) \end{gathered}$$$

Differentiability at $x=1$

$$$\begin{gathered} {f}^{'}(x)= 2\alpha (2x)+2\beta , x1 \ = \ {f}^{'}({1}^{-})={f}^{'}({1}^{+}) \ 2\alpha (2\cdot 1)+2\beta =\alpha +3 \ 3\alpha +2\beta =3-(2) \end{gathered}$$$

Solving equation (1) and (2)

$$$\begin{gathered} (1)\times 2-(2)\times 3 \ -8\alpha -9\alpha =-3\implies \alpha =\frac{3}{17} \end{gathered}$$$

Substitute $\alpha =\frac{3}{17}$ in equation (1):

$$$\begin{gathered} 3\beta -4\times \frac{3}{17}=3 \ 3\beta =3+\frac{12}{17}=\frac{63}{17} \end{gathered}$$$

$\beta =\frac{21}{17}$

Value of $

Solutions · 1