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Let-f-x-3x-2-1-x-lt-0-cx-d-0-x-1-x-8-x-gt-1-find-the-value-of-c-amp-d-such-that-f-x-continous-everywhere-




Question Number 140073 by EDWIN88 last updated on 04/May/21
Let f(x)= { ((3x^2 −1 ; x<0)),((cx+d ; 0≤x≤1)),(((√(x+8)) ; x>1)) :}  find the value of c & d such that f(x) continous  everywhere
$$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{3x}^{\mathrm{2}} −\mathrm{1}\:;\:\mathrm{x}<\mathrm{0}}\\{\mathrm{cx}+\mathrm{d}\:;\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}}\\{\sqrt{\mathrm{x}+\mathrm{8}}\:;\:\mathrm{x}>\mathrm{1}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{c}\:\&\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{continous} \\ $$$$\mathrm{everywhere} \\ $$
Answered by bobhans last updated on 04/May/21
since lim_(x→0^( −) )  3x^2 −1 =−1 the value of  cx+d at x=0 must be −1 that is d=−1  since lim_(x→1^( +) )  (√(x+8)) = 3 the value of cx+d at x=1  must be 3 that is 3=c(1)−1, c=4  ∴  { ((c=4)),((d=−1)) :}
$$\mathrm{since}\:\underset{{x}\rightarrow\mathrm{0}^{\:−} } {\mathrm{lim}}\:\mathrm{3x}^{\mathrm{2}} −\mathrm{1}\:=−\mathrm{1}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cx}+\mathrm{d}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:\mathrm{must}\:\mathrm{be}\:−\mathrm{1}\:\mathrm{that}\:\mathrm{is}\:\mathrm{d}=−\mathrm{1} \\ $$$$\mathrm{since}\:\underset{{x}\rightarrow\mathrm{1}^{\:+} } {\mathrm{lim}}\:\sqrt{\mathrm{x}+\mathrm{8}}\:=\:\mathrm{3}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cx}+\mathrm{d}\:\mathrm{at}\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{3}\:\mathrm{that}\:\mathrm{is}\:\mathrm{3}=\mathrm{c}\left(\mathrm{1}\right)−\mathrm{1},\:\mathrm{c}=\mathrm{4} \\ $$$$\therefore\:\begin{cases}{\mathrm{c}=\mathrm{4}}\\{\mathrm{d}=−\mathrm{1}}\end{cases} \\ $$

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