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Question Number 140073 by EDWIN88 last updated on 04/May/21

$$\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}{3\mathrm{x}}^2-1;\mathrm{x}<0\\ \mathrm{cx}+\mathrm{d};0⩽\mathrm{x}⩽1\\ \sqrt{\mathrm{x}+8};\mathrm{x}>1\end{array}$$ $$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{c}\mathrm{\&}\mathrm{d}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)\mathrm{continous}$$ $$\mathrm{everywhere}$$

Answered by bobhans last updated on 04/May/21

$$\mathrm{since}\underset{x\to 0^{-}}{\lim }{3\mathrm{x}}^2-1=-1\mathrm{the}\mathrm{value}\mathrm{of}$$ $$\mathrm{cx}+\mathrm{d}\mathrm{at}\mathrm{x}=0\mathrm{must}\mathrm{be}-1\mathrm{that}\mathrm{is}\mathrm{d}=-1$$ $$\mathrm{since}\underset{x\to 1^{+}}{\lim }\sqrt{\mathrm{x}+8}=3\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{cx}+\mathrm{d}\mathrm{at}\mathrm{x}=1$$ $$\mathrm{must}\mathrm{be}3\mathrm{that}\mathrm{is}3=\mathrm{c}\left(1\right)-1,\mathrm{c}=4$$ $$\therefore \{\begin{array}{l}\mathrm{c}=4\\ \mathrm{d}=-1\end{array}$$

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