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Question Number 52542 by hassentimol last updated on 09/Jan/19

Let f(x) = (( 5x + 40 )/(x + 2)) (1) Calculate the derivative of f(4) without using the answer of question (2), (2) Calculate the derivative of f(x), (3) Calculate lim_(x → +∞) f(x) . All your answers should be correctly proved and detailed. Can you please help me for that exercise.

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\frac{\:\mathrm{5}{x}\:+\:\mathrm{40}\:}{{x}\:+\:\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{f}\left(\mathrm{4}\right)\:\mathrm{without} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{question}\:\left(\mathrm{2}\right), \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{f}\left({x}\right), \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Calculate}\:\:\underset{{x}\:\rightarrow\:+\infty} {\mathrm{lim}}\:{f}\left({x}\right)\:. \\ $$$$\mathrm{All}\:\mathrm{your}\:\mathrm{answers}\:\mathrm{should}\:\mathrm{be}\:\mathrm{correctly}\:\mathrm{proved}\:\mathrm{and}\:\mathrm{detailed}. \\ $$$$ \\ $$$${Can}\:{you}\:{please}\:{help}\:{me}\:{for}\:{that}\:{exercise}. \\ $$

Commented by Abdo msup. last updated on 10/Jan/19

1)f^′ (4)=lim_(h→0) ((f(4+h)−f(4))/h) =lim_(h→0) ((((5(4+h)+40)/(4+h+2))−10)/h) =lim_(h→0) ((60+5h−60−10h)/(h(6+h))) =lim_(h→0) ((−5h)/(h(6+h))) =lim_(h→0) ((−5)/(6+h)) =−(5/6) 2)f(x)=((5(x+2)+30)/(x+2)) =5+((30)/(x+2))) ⇒ f^′ (x)=−((30)/((x+2)^2 )) for x≠2 .

$$\left.\mathrm{1}\right){f}^{'} \left(\mathrm{4}\right)={lim}_{{h}\rightarrow\mathrm{0}} \frac{{f}\left(\mathrm{4}+{h}\right)−{f}\left(\mathrm{4}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\frac{\frac{\mathrm{5}\left(\mathrm{4}+{h}\right)+\mathrm{40}}{\mathrm{4}+{h}+\mathrm{2}}−\mathrm{10}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{60}+\mathrm{5}{h}−\mathrm{60}−\mathrm{10}{h}}{{h}\left(\mathrm{6}+{h}\right)} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \frac{−\mathrm{5}{h}}{{h}\left(\mathrm{6}+{h}\right)}\:={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{−\mathrm{5}}{\mathrm{6}+{h}}\:=−\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)=\frac{\mathrm{5}\left({x}+\mathrm{2}\right)+\mathrm{30}}{{x}+\mathrm{2}}\:=\mathrm{5}+\frac{\mathrm{30}}{\left.{x}+\mathrm{2}\right)}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=−\frac{\mathrm{30}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:{for}\:{x}\neq\mathrm{2}\:. \\ $$

Commented by Abdo msup. last updated on 10/Jan/19

3)lim_(x→+∞) f(x)=lim_(x→+∞) ((5x)/x) =5 .

$$\left.\mathrm{3}\right){lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow+\infty} \frac{\mathrm{5}{x}}{{x}}\:=\mathrm{5}\:. \\ $$

Commented by hassentimol last updated on 10/Jan/19

Thanks sir !!

$$\mathrm{Thanks}\:\mathrm{sir}\:!! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

2)((df(x))/dx)=(((x+2)((d(5x+40))/dx)−(5x+40)(d/dx)(x+2))/((x+2)^2 )) =(((x+2)(5)−(5x+40)(1))/((x+2)^2 )) =((5x+10−5x−40)/((x+2)^2 )) =((−30)/((x+2)^2 )) 3)lim_(x→∞) ((5x+40)/(x+2)) =lim_(x→∞) ((5+((40)/x))/(1+(2/x))) =((5+0)/(1+0))=5 1)f(x)=((5x+40)/(x+2)) ((df/dx))_(at x=4) = lim_(h→0) ((f(4+h)−f(4))/h) =lim_(h→0) ((((5(4+h)+40)/((4+h)+2))−((60)/6))/h) =lim_(h→0) ((20+5h+40−40−10h−20)/(h(6+h)6)) =lim_(h→0) ((−5h)/(h(36+6h)))=((−5)/(36))

$$\left.\mathrm{2}\right)\frac{{df}\left({x}\right)}{{dx}}=\frac{\left({x}+\mathrm{2}\right)\frac{{d}\left(\mathrm{5}{x}+\mathrm{40}\right)}{{dx}}−\left(\mathrm{5}{x}+\mathrm{40}\right)\frac{{d}}{{dx}}\left({x}+\mathrm{2}\right)}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{5}\right)−\left(\mathrm{5}{x}+\mathrm{40}\right)\left(\mathrm{1}\right)}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}{x}+\mathrm{10}−\mathrm{5}{x}−\mathrm{40}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{30}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{5}{x}+\mathrm{40}}{{x}+\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{5}+\frac{\mathrm{40}}{{x}}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}} \\ $$$$=\frac{\mathrm{5}+\mathrm{0}}{\mathrm{1}+\mathrm{0}}=\mathrm{5} \\ $$$$\left.\mathrm{1}\right){f}\left({x}\right)=\frac{\mathrm{5}{x}+\mathrm{40}}{{x}+\mathrm{2}} \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{at}\:{x}=\mathrm{4}} =\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left(\mathrm{4}+{h}\right)−{f}\left(\mathrm{4}\right)}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{5}\left(\mathrm{4}+{h}\right)+\mathrm{40}}{\left(\mathrm{4}+{h}\right)+\mathrm{2}}−\frac{\mathrm{60}}{\mathrm{6}}}{{h}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{20}+\mathrm{5}{h}+\mathrm{40}−\mathrm{40}−\mathrm{10}{h}−\mathrm{20}}{{h}\left(\mathrm{6}+{h}\right)\mathrm{6}} \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{5}{h}}{{h}\left(\mathrm{36}+\mathrm{6}{h}\right)}=\frac{−\mathrm{5}}{\mathrm{36}} \\ $$

Commented by hassentimol last updated on 09/Jan/19

Thank you sir ! God bless you Thank you very much, I′m now understanding.

$${Thank}\:{you}\:{sir}\:!\:{God}\:{bless}\:{you} \\ $$$${Thank}\:{you}\:{very}\:{much},\:{I}'{m}\:{now}\:{understanding}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

most welcome...

$${most}\:{welcome}... \\ $$

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