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Question Number 120721 by help last updated on 02/Nov/20

Commented by JDamian last updated on 02/Nov/20

the first range (x>q), is it correct?

$${the}\:{first}\:{range}\:\left({x}>{q}\right),\:{is}\:{it}\:{correct}? \\ $$

Commented by help last updated on 02/Nov/20

x<q (correction)

$${x}<{q}\:\left({correction}\right) \\ $$

Answered by john santu last updated on 02/Nov/20

(1)f(2)=8−p=p ⇒ p=4 (2) lim_(x→0) f(x)=lim_(x→0) f(x) lim_(x→0) r(x+6)=lim_(x→0) (((2−x)^2 −p)/x) , it follows that 6r = lim_(x→0) (((2−x+2)(2−x−2))/x) 6r = lim_(x→0) (((4−x)(−x))/x) ; 6r=−4 ⇒r=−(2/3) then q=0

$$\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)=\mathrm{8}−{p}={p}\:\Rightarrow\:{p}=\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right) \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{r}\left({x}+\mathrm{6}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} −{p}}{{x}}\:,\:{it}\:{follows}\: \\ $$$${that}\:\mathrm{6}{r}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}−{x}+\mathrm{2}\right)\left(\mathrm{2}−{x}−\mathrm{2}\right)}{{x}} \\ $$$$\:\mathrm{6}{r}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{4}−{x}\right)\left(−{x}\right)}{{x}}\:;\:\mathrm{6}{r}=−\mathrm{4}\:\Rightarrow{r}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${then}\:{q}=\mathrm{0} \\ $$

Commented by JDamian last updated on 02/Nov/20

I think that r(x+6)∣_(x=2) = f(2)=p 8r=p r=(p/8)=(1/2)

$${I}\:{think}\:{that}\:\:\:{r}\left({x}+\mathrm{6}\right)\mid_{{x}=\mathrm{2}} =\:{f}\left(\mathrm{2}\right)={p} \\ $$$$\mathrm{8}{r}={p} \\ $$$${r}=\frac{{p}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by help last updated on 02/Nov/20

thx pls check the answer for r

$${thx}\:{pls}\:{check}\:{the}\:{answer}\:{for}\:{r} \\ $$

Answered by liberty last updated on 02/Nov/20

f(x)= { (((((2−x)^2 −p)/x) ; x<q)),((r(x+6) ; q≤x<2 )),((x^3 −p ; x≥2)) :} Find the values of p,q and r such that f(x) is continous everywhere and f(2)=p. (i) f(2)= 8−p=p ⇒p=4 (ii) lim_(x→2) f(x)=f(2) ; lim_(x→2) r(x+6)=4 8r = 4 ⇒r=(1/2) (iii) lim_(x→q) f(x) = lim_(x→q) f(x)⇒lim_(x→q) ((4−4x+x^2 −4)/x) = lim_(x→q) (1/2)(x+6) lim_(x→q) = ((x(x−4))/x) = ((q+6)/2) ; q−4 = ((q+6)/2) 2q−8 = q+6 ⇒ q=14

$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\frac{\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{p}}{\mathrm{x}}\:;\:\mathrm{x}<\mathrm{q}}\\{\mathrm{r}\left(\mathrm{x}+\mathrm{6}\right)\:;\:\mathrm{q}\leqslant\mathrm{x}<\mathrm{2}\:}\\{\mathrm{x}^{\mathrm{3}} −\mathrm{p}\:;\:\mathrm{x}\geqslant\mathrm{2}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$$$\mathrm{continous}\:\mathrm{everywhere}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{p}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{2}\right)=\:\mathrm{8}−\mathrm{p}=\mathrm{p}\:\Rightarrow\mathrm{p}=\mathrm{4} \\ $$$$\left(\mathrm{ii}\right)\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{2}\right)\:;\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{r}\left(\mathrm{x}+\mathrm{6}\right)=\mathrm{4} \\ $$$$\:\:\mathrm{8r}\:=\:\mathrm{4}\:\Rightarrow\mathrm{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{iii}\right)\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\Rightarrow\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\frac{\mathrm{4}−\mathrm{4x}+\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+\mathrm{6}\right) \\ $$$$\:\underset{{x}\rightarrow\mathrm{q}} {\mathrm{lim}}\:=\:\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right)}{\mathrm{x}}\:=\:\frac{\mathrm{q}+\mathrm{6}}{\mathrm{2}}\:;\:\mathrm{q}−\mathrm{4}\:=\:\frac{\mathrm{q}+\mathrm{6}}{\mathrm{2}} \\ $$$$\:\mathrm{2q}−\mathrm{8}\:=\:\mathrm{q}+\mathrm{6}\:\Rightarrow\:\mathrm{q}=\mathrm{14} \\ $$

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