If $\begin{aligned} f(x) &=6 \beta-3 \propto x, ext { if }-4 \leq x

Question

If $\begin{aligned} f(x) &=6 \beta-3 \propto x, 	ext { if }-4 \leq x < -2 \ &=4 x+1, 	ext { if }-2 \leq x \leq 2 \end{aligned}$ is continuous on $[-4,2]$, then $\propto+\beta=$

Aicharya · Accepted Answer

$\lim _{x ightarrow-2^{-}} f(x)=\lim _{x ightarrow-2^{-}}(6 \beta-3 \alpha x)$ $=6 \beta+6 \alpha$ ...(1) $\lim _{x ightarrow-2^{+}} f(x)=\lim _{x ightarrow-2^{+}}(4 x+1)=-7$ ...(2) Given $f(x)$ is continuous on $[-4,2]$ $	herefore 6 \beta+6 \alpha=-7$ $	herefore \alpha+\beta=\frac{-7}{6}$

If $\begin{aligned} f(x) &=6 \beta-3 \propto x, \text { if }-4 \leq x < -2 \\ &=4 x+1, \text { if }-2 \leq x…

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