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

Commented by JDamian last updated on 02/Nov/20

$$thefirstrange(x>q),isitcorrect?$$

Commented by help last updated on 02/Nov/20

$$x<q\left(correction\right)$$

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

$$\left(1\right)f\left(2\right)=8-p=p\Rightarrow p=4$$$$\left(2\right)\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }f\left(x\right)$$$$\underset{x\to 0}{\lim }r(x+6)=\underset{x\to 0}{\lim }\frac{{(2-x)}^2-p}{x},itfollows$$$$that6r=\underset{x\to 0}{\lim }\frac{(2-x+2)(2-x-2)}{x}$$$$6r=\underset{x\to 0}{\lim }\frac{(4-x)(-x)}{x};6r=-4\Rightarrow r=-\frac{2}{3}$$$$thenq=0$$

Commented by JDamian last updated on 02/Nov/20

$$Ithinkthatr(x+6){\mid }_{x=2}=f\left(2\right)=p$$$$8r=p$$$$r=\frac{p}{8}=\frac{1}{2}$$

Commented by help last updated on 02/Nov/20

$$thxplschecktheanswerforr$$

Answered by liberty last updated on 02/Nov/20

$$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}\frac{{(2-\mathrm{x})}^2-\mathrm{p}}{\mathrm{x}};\mathrm{x}<\mathrm{q}\\ \mathrm{r}(\mathrm{x}+6);\mathrm{q}⩽\mathrm{x}<2\\ {\mathrm{x}}^3-\mathrm{p};\mathrm{x}⩾2\end{array}$$$$\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{p},\mathrm{q}\mathrm{and}\mathrm{r}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}$$$$\mathrm{continous}\mathrm{everywhere}\mathrm{and}\mathrm{f}\left(2\right)=\mathrm{p}.$$$$\left(\mathrm{i}\right)\mathrm{f}\left(2\right)=8-\mathrm{p}=\mathrm{p}\Rightarrow \mathrm{p}=4$$$$\left(\mathrm{ii}\right)\underset{x\to 2}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(2\right);\underset{x\to 2}{\lim }\mathrm{r}(\mathrm{x}+6)=4$$$$8\mathrm{r}=4\Rightarrow \mathrm{r}=\frac{1}{2}$$$$\left(\mathrm{iii}\right)\underset{x\to \mathrm{q}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{x\to \mathrm{q}}{\lim }\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \underset{x\to \mathrm{q}}{\lim }\frac{4-4\mathrm{x}+{\mathrm{x}}^2-4}{\mathrm{x}}=\underset{x\to \mathrm{q}}{\lim }\frac{1}{2}(\mathrm{x}+6)$$$$\underset{x\to \mathrm{q}}{\lim }=\frac{\mathrm{x}(\mathrm{x}-4)}{\mathrm{x}}=\frac{\mathrm{q}+6}{2};\mathrm{q}-4=\frac{\mathrm{q}+6}{2}$$$$2\mathrm{q}-8=\mathrm{q}+6\Rightarrow \mathrm{q}=14$$

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