Question Number 120721 by help last updated on 02/Nov/20
Commented by JDamian last updated on 02/Nov/20
$${the}\:{first}\:{range}\:\left({x}>{q}\right),\:{is}\:{it}\:{correct}? \\ $$
Commented by help last updated on 02/Nov/20
$${x}<{q}\:\left({correction}\right) \\ $$
Answered by john santu last updated on 02/Nov/20
$$\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)=\mathrm{8}−{p}={p}\:\Rightarrow\:{p}=\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right) \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{r}\left({x}+\mathrm{6}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} −{p}}{{x}}\:,\:{it}\:{follows}\: \\ $$$${that}\:\mathrm{6}{r}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}−{x}+\mathrm{2}\right)\left(\mathrm{2}−{x}−\mathrm{2}\right)}{{x}} \\ $$$$\:\mathrm{6}{r}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{4}−{x}\right)\left(−{x}\right)}{{x}}\:;\:\mathrm{6}{r}=−\mathrm{4}\:\Rightarrow{r}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${then}\:{q}=\mathrm{0} \\ $$
Commented by JDamian last updated on 02/Nov/20
$${I}\:{think}\:{that}\:\:\:{r}\left({x}+\mathrm{6}\right)\mid_{{x}=\mathrm{2}} =\:{f}\left(\mathrm{2}\right)={p} \\ $$$$\mathrm{8}{r}={p} \\ $$$${r}=\frac{{p}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by help last updated on 02/Nov/20
$${thx}\:{pls}\:{check}\:{the}\:{answer}\:{for}\:{r} \\ $$
Answered by liberty last updated on 02/Nov/20
$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\frac{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{p}}{\mathrm{x}}\:;\:\mathrm{x}<\mathrm{q}}\\{\mathrm{r}\left(\mathrm{x}+\mathrm{6}\right)\:;\:\mathrm{q}\leqslant\mathrm{x}<\mathrm{2}\:}\\{\mathrm{x}^{\mathrm{3}} −\mathrm{p}\:;\:\mathrm{x}\geqslant\mathrm{2}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$$$\mathrm{continous}\:\mathrm{everywhere}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{p}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{2}\right)=\:\mathrm{8}−\mathrm{p}=\mathrm{p}\:\Rightarrow\mathrm{p}=\mathrm{4} \\ $$$$\left(\mathrm{ii}\right)\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{2}\right)\:;\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{r}\left(\mathrm{x}+\mathrm{6}\right)=\mathrm{4} \\ $$$$\:\:\mathrm{8r}\:=\:\mathrm{4}\:\Rightarrow\mathrm{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{iii}\right)\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\Rightarrow\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\frac{\mathrm{4}−\mathrm{4x}+\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+\mathrm{6}\right) \\ $$$$\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:=\:\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right)}{\mathrm{x}}\:=\:\frac{\mathrm{q}+\mathrm{6}}{\mathrm{2}}\:;\:\mathrm{q}−\mathrm{4}\:=\:\frac{\mathrm{q}+\mathrm{6}}{\mathrm{2}} \\ $$$$\:\mathrm{2q}−\mathrm{8}\:=\:\mathrm{q}+\mathrm{6}\:\Rightarrow\:\mathrm{q}=\mathrm{14} \\ $$