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Question Number 80180 by mathocean1 last updated on 31/Jan/20

h(x)=((x−x^2 )/(x+1)) we defined this function on R−{−1}→R 1) Study the variations of h then draw up its table of variation. please sirs i need your kind help

$$\mathrm{h}\left({x}\right)=\frac{{x}−{x}^{\mathrm{2}} }{{x}+\mathrm{1}} \\ $$$${we}\:{defined}\:{this}\:{function}\:{on} \\ $$$$\mathbb{R}−\left\{−\mathrm{1}\right\}\rightarrow\mathbb{R} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Study}\:\mathrm{the}\:\mathrm{variations}\:\mathrm{of}\:\mathrm{h}\:\mathrm{then} \\ $$$$\mathrm{draw}\:\mathrm{up}\:\mathrm{its}\:\mathrm{table}\:\mathrm{of}\:\mathrm{variation}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{sirs}\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{kind}\:\mathrm{help} \\ $$

Answered by Rio Michael last updated on 31/Jan/20

solution −−−−−−−−−− h(x) = ((x−x^2 )/(x−1)) h(x) = 0 ⇒ x −x^2 = 0 ⇒ x =0 or x = 1 h′(x) = (((x−x^2 )−(x−1)(1−2x))/((x−1)^2 )) = ((x −x^2 −(x−2x^2 −1+2x))/((x−1)^2 )) h^′ (x) = ((x^2 −2x + 1)/((x−1)^2 )) h^′ (x) > 0 ⇒ (x−1)^2 > 0 ⇒ x > 1 h is increasing for x > 1 and decreasing other wise

$$\:\boldsymbol{\mathrm{solution}} \\ $$$$\:−−−−−−−−−− \\ $$$$\mathrm{h}\left({x}\right)\:=\:\frac{{x}−{x}^{\mathrm{2}} }{{x}−\mathrm{1}} \\ $$$${h}\left({x}\right)\:=\:\mathrm{0}\:\Rightarrow\:{x}\:−{x}^{\mathrm{2}} \:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\mathrm{0}\:{or}\:{x}\:=\:\mathrm{1} \\ $$$$\:\mathrm{h}'\left({x}\right)\:=\:\frac{\left({x}−{x}^{\mathrm{2}} \right)−\left({x}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{{x}\:−{x}^{\mathrm{2}} \:−\left({x}−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:{h}^{'} \left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\:\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\:{h}^{'} \left({x}\right)\:>\:\mathrm{0}\:\Rightarrow\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:>\:\mathrm{0}\:\:\Rightarrow\:{x}\:>\:\mathrm{1} \\ $$$$\:{h}\:{is}\:{increasing}\:{for}\:{x}\:>\:\mathrm{1}\:{and}\:{decreasing}\:{other}\:{wise} \\ $$$$ \\ $$

Commented by mathocean1 last updated on 31/Jan/20

Please sir h′ is not like this: h′=(((1−2x)(x+1)−(x−x^2 ))/((x+1)^2 )) ?

$$\mathrm{Please}\:\mathrm{sir}\:\mathrm{h}'\:\mathrm{is}\:\mathrm{not}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\mathrm{h}'=\frac{\left(\mathrm{1}−\mathrm{2}{x}\right)\left({x}+\mathrm{1}\right)−\left({x}−{x}^{\mathrm{2}} \right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:?\:\: \\ $$$$\: \\ $$

Commented by mathocean1 last updated on 31/Jan/20

h(x)=((x−x^2 )/(x+1))

$$\mathrm{h}\left({x}\right)=\frac{{x}−{x}^{\mathrm{2}} }{{x}+\mathrm{1}} \\ $$

Commented by Rio Michael last updated on 31/Jan/20

sorry i copied the question wrongly

$${sorry}\:{i}\:{copied}\:{the}\:{question}\:{wrongly} \\ $$

Commented by mathocean1 last updated on 31/Jan/20

please sir how can i draw up its table of variation when h′=((−x^2 −2x+1)/((x+1)^2 ))=−1 ?

$$\mathrm{please}\:\mathrm{sir}\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{draw}\:\mathrm{up}\:\mathrm{its}\: \\ $$$$\mathrm{table}\:\mathrm{of}\:\mathrm{variation}\:\mathrm{when}\: \\ $$$$\mathrm{h}'=\frac{−{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=−\mathrm{1}\:? \\ $$

Commented by JDamian last updated on 01/Feb/20

if h(x) were ((x−x^2 )/(x−1)), then h(x)=((x(1−x))/(x−1))=−x a very easy function

$${if}\:\:\mathrm{h}\left({x}\right)\:{were}\:\frac{{x}−{x}^{\mathrm{2}} }{{x}−\mathrm{1}},\:{then}\:\mathrm{h}\left({x}\right)=\frac{{x}\left(\mathrm{1}−{x}\right)}{{x}−\mathrm{1}}=−{x} \\ $$$${a}\:{very}\:{easy}\:{function} \\ $$

Commented by mathocean1 last updated on 01/Feb/20

so... we can study h(x)=((x−x^2 )/(x+1))???

$$\mathrm{so}...\:\mathrm{we}\:\mathrm{can}\:\mathrm{study}\:\mathrm{h}\left({x}\right)=\frac{{x}−{x}^{\mathrm{2}} }{{x}+\mathrm{1}}??? \\ $$

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