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Question Number 78628 by naka3546 last updated on 19/Jan/20
Find minimum value of y = ((2x)/(x^2 + x + 1)) x , y ∈ R Without Differential
$${Find}\:\:{minimum}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\:{y}\:\:=\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}} \\ $$$${x}\:,\:{y}\:\:\in\:\:\mathbb{R} \\ $$$${Without}\:\:{Differential} \\ $$
Commented by john santu last updated on 19/Jan/20
denominator x^2 +x+1 >0 ∀x∈R x^2 y+xy+y−2x=0 x^2 y+(y−2)x+y=0 △= (y−2)^2 −4y.y≥0 (y−2)^2 −(2y)^2 ≥0 (3y−2)(−y−2)≥0 (3y−2)(y+2)≤0 −2≤y≤(2/3) { ((minimum = −2)),((maximum = (2/3))) :}
$${denominator}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:>\mathrm{0}\:\forall{x}\in\mathbb{R} \\ $$$${x}^{\mathrm{2}} {y}+{xy}+{y}−\mathrm{2}{x}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} {y}+\left({y}−\mathrm{2}\right){x}+{y}=\mathrm{0} \\ $$$$\bigtriangleup=\:\left({y}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{y}.{y}\geqslant\mathrm{0} \\ $$$$\left({y}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}{y}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{y}−\mathrm{2}\right)\left(−{y}−\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{y}−\mathrm{2}\right)\left({y}+\mathrm{2}\right)\leqslant\mathrm{0} \\ $$$$−\mathrm{2}\leqslant{y}\leqslant\frac{\mathrm{2}}{\mathrm{3}}\:\begin{cases}{{minimum}\:=\:−\mathrm{2}}\\{{maximum}\:=\:\frac{\mathrm{2}}{\mathrm{3}}}\end{cases} \\ $$
Answered by jagoll last updated on 19/Jan/20
lim_(x→∞) ((2x)/(x^2 +x+1)) = 0 lim_(x→−∞) ((2x)/(x^2 +x+1)) =0 y_(min) = 0
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\mathrm{0}\: \\ $$$$\mathrm{y}_{\mathrm{min}} \:=\:\mathrm{0} \\ $$
Commented by mr W last updated on 19/Jan/20
what do you get when you put x=−1?
$${what}\:{do}\:{you}\:{get}\:{when}\:{you}\:{put}\:{x}=−\mathrm{1}? \\ $$
Commented by jagoll last updated on 19/Jan/20
x=−1 ⇒y = ((−2)/1) =−2 sir?
$$\mathrm{x}=−\mathrm{1}\:\Rightarrow\mathrm{y}\:=\:\frac{−\mathrm{2}}{\mathrm{1}}\:=−\mathrm{2}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 19/Jan/20
well ... what minimum value sir?
$$\mathrm{well}\:…\:\mathrm{what}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 19/Jan/20
if x>0: y=(2/(1+x+(1/x)))=(2/(3+((√x)−(1/( (√x))))^2 )) if x<0: y=(2/(1+x+(1/x)))=(2/(−1−((√(−x))−(1/( (√(−x)))))^2 )) y(0)=0 and lim_(x→+∞) y=0, y(x>0)>0 ⇒there is a maximum for x>0 it is to see when (√x)=(1/( (√x))), i.e. x=1, y_(max) =(2/3) y(0)=0 and lim_(x→−∞) y=0, y(x<0)<0 ⇒there is a minimum for x<0 it is to see when (√(−x))=(1/( (√(−x)))), i.e. x=−1, y_(min) =(2/(−1))=−2
$${if}\:{x}>\mathrm{0}:\:{y}=\frac{\mathrm{2}}{\mathrm{1}+{x}+\frac{\mathrm{1}}{{x}}}=\frac{\mathrm{2}}{\mathrm{3}+\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} } \\ $$$${if}\:{x}<\mathrm{0}:\:{y}=\frac{\mathrm{2}}{\mathrm{1}+{x}+\frac{\mathrm{1}}{{x}}}=\frac{\mathrm{2}}{−\mathrm{1}−\left(\sqrt{−{x}}−\frac{\mathrm{1}}{\:\sqrt{−{x}}}\right)^{\mathrm{2}} } \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{y}=\mathrm{0},\:{y}\left({x}>\mathrm{0}\right)>\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{a}\:{maximum}\:{for}\:{x}>\mathrm{0} \\ $$$${it}\:{is}\:{to}\:{see}\:{when}\:\sqrt{{x}}=\frac{\mathrm{1}}{\:\sqrt{{x}}},\:{i}.{e}.\:{x}=\mathrm{1},\:{y}_{{max}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{y}=\mathrm{0},\:{y}\left({x}<\mathrm{0}\right)<\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{a}\:{minimum}\:{for}\:{x}<\mathrm{0} \\ $$$${it}\:{is}\:{to}\:{see}\:{when}\:\sqrt{−{x}}=\frac{\mathrm{1}}{\:\sqrt{−{x}}},\:{i}.{e}.\:{x}=−\mathrm{1},\:{y}_{{min}} =\frac{\mathrm{2}}{−\mathrm{1}}=−\mathrm{2} \\ $$
Commented by john santu last updated on 19/Jan/20
the last line y _(min ) or y_(max ) ?
$${the}\:{last}\:{line}\:{y}\:_{{min}\:} \:{or}\:{y}_{{max}\:} ? \\ $$
Commented by mr W last updated on 19/Jan/20
typo. y_(min) should be.
$${typo}.\:{y}_{{min}} \:{should}\:{be}. \\ $$
Answered by behi83417@gmail.com last updated on 19/Jan/20
(1/y)=((x^2 +x+1)/(2x))=(1/2)(x+(1/x)+1)≥(1/2)(2+1)=(3/2) ⇒y≤(2/3)
$$\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{2x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}+\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}\leqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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