Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\mathrm{Let}f\left(x\right)=\frac{5x+40}{x+2}$$$$\left(1\right)\mathrm{Calculate}\mathrm{the}\mathrm{derivative}\mathrm{of}f\left(4\right)\mathrm{without}$$$$\mathrm{using}\mathrm{the}\mathrm{answer}\mathrm{of}\mathrm{question}\left(2\right),$$$$\left(2\right)\mathrm{Calculate}\mathrm{the}\mathrm{derivative}\mathrm{of}f\left(x\right),$$$$\left(3\right)\mathrm{Calculate}\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right).$$$$\mathrm{All}\mathrm{your}\mathrm{answers}\mathrm{should}\mathrm{be}\mathrm{correctly}\mathrm{proved}\mathrm{and}\mathrm{detailed}.$$$$$$$$Canyoupleasehelpmeforthatexercise.$$

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

$$1)f^{{}^{\prime }}\left(4\right)={lim}_{h\to 0}\frac{f(4+h)-f\left(4\right)}{h}$$$$={lim}_{h\to 0}\frac{\frac{5(4+h)+40}{4+h+2}-10}{h}$$$$={lim}_{h\to 0}\frac{60+5h-60-10h}{h(6+h)}$$$$={lim}_{h\to 0}\frac{-5h}{h(6+h)}={lim}_{h\to 0}\frac{-5}{6+h}=-\frac{5}{6}$$$$2)f\left(x\right)=\frac{5(x+2)+30}{x+2}=5+\frac{30}{x+2)}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=-\frac{30}{{(x+2)}^2}forx\ne 2.$$

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

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

Commented by hassentimol last updated on 10/Jan/19

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

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

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

Commented by hassentimol last updated on 09/Jan/19

$$Thankyousir!Godblessyou$$$$Thankyouverymuch,I^{\prime }mnowunderstanding.$$

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

$$mostwelcome...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com