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If-x-R-the-least-value-of-the-expression-x-2-6x-5-x-2-2x-1-is-

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Question Number 81672 by zainal tanjung last updated on 14/Feb/20
If x ∈ R, the least value of the expression ((x^2 −6x+5)/(x^2 +2x+1)) is
$$\mathrm{If}\:{x}\:\in\:{R},\:\mathrm{the}\:\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{expression}\:\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}\:\mathrm{is} \\ $$
Commented by Kunal12588 last updated on 14/Feb/20
y=(((x−5)(x−1))/((x+1)^2 )) (dy/dx)=(((x+1)^2 [(x−5)+(x−1)]−(x−5)(x−1)[2(x+1)])/((x+1)^4 )) ⇒(dy/dx)=((2(x^2 +2x+1)(x−3)−2(x−5)(x^2 −1))/((x+1)^4 )) ⇒(dy/dx)=((2[x^3 +2x^2 +x−3x^2 −6x−3−(x^3 −5x^2 −x+5)])/((x+1)^4 )) ⇒(dy/dx)=((2(4x^2 −4x−8))/((x+1)^4 )) ⇒(dy/dx)=((8(x^2 −x−2))/((x+1)^4 )) ⇒(dy/dx)=((8(x−2)(x+1))/((x+1)^4 )) for max or min (dy/dx)=0⇒x=2, x≠−1 (d^2 y/dx^2 )>0 when x=2 y_(local min) =(((2−5)(2−1))/((2+1)^2 ))=((−3)/3^2 )=−(1/3)=−0.3^(−)
$${y}=\frac{\left({x}−\mathrm{5}\right)\left({x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left[\left({x}−\mathrm{5}\right)+\left({x}−\mathrm{1}\right)\right]−\left({x}−\mathrm{5}\right)\left({x}−\mathrm{1}\right)\left[\mathrm{2}\left({x}+\mathrm{1}\right)\right]}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)−\mathrm{2}\left({x}−\mathrm{5}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{2}\left[{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{3}−\left({x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{5}\right)\right]}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{8}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{8}\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{8}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$${for}\:{max}\:{or}\:{min} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{0}\Rightarrow{x}=\mathrm{2},\:{x}\neq−\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }>\mathrm{0}\:{when}\:{x}=\mathrm{2} \\ $$$${y}_{{local}\:{min}} =\frac{\left(\mathrm{2}−\mathrm{5}\right)\left(\mathrm{2}−\mathrm{1}\right)}{\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{−\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}}=−\mathrm{0}.\overline {\mathrm{3}} \\ $$
Commented by Tony Lin last updated on 14/Feb/20
let ((x^2 −6x+5)/(x^2 +2x+1))=k ⇒x^2 −6x+5=kx^2 +2kx+k ⇒(k−1)x^2 +(2k+6)x+(k−5)=0 Δ=(2k+6)^2 −4(k−1)(k−5)=0 ⇒k=−(1/3) when x=2 ((x^2 −6x+5)/(x^2 +2x+1))=−(1/3)min
$${let}\:\:\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}={k} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}={kx}^{\mathrm{2}} +\mathrm{2}{kx}+{k} \\ $$$$\Rightarrow\left({k}−\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{6}\right){x}+\left({k}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}{k}+\mathrm{6}\right)^{\mathrm{2}} −\mathrm{4}\left({k}−\mathrm{1}\right)\left({k}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=−\frac{\mathrm{1}}{\mathrm{3}}\:{when}\:{x}=\mathrm{2} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{3}}{min} \\ $$
Commented by mathmax by abdo last updated on 14/Feb/20
f(x)=((x^2 +2x+1−8x+4)/(x^2 +2x+1)) =1+((4−8x)/(x^2 +2x+1)) ⇒ f^′ (x)=−4(((2x−1)/(x^2 +2x+1)))^((1)) =−4(((2(x^2 +2x+1)−(2x−1)(2x+2))/((x^2 +2x+1)^2 ))) =−4×((2x^2 +4x+2−4x^2 −4x+2x+2)/((x+1)^4 ))=−4×((−2x^2 +2x+4)/((x+1)^4 )) =8×((x^2 −x−2)/((x+1)^4 )) so f^′ (x)=0 ⇔x^2 −x−2=0 and x≠−1 Δ=1−4(−2)=9 ⇒x_1 =((1+3)/2)=2 and x_2 =((1−3)/2)=−1 x −∞ −1 2 +∞ f^′ + ∣∣ − 0 + f incr decr f(2) incr inf f(x) =f(2)=((4−12+5)/(4+4+1)) =((−3)/9) =−(1/3)
$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{8}{x}+\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{4}−\mathrm{8}{x}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=−\mathrm{4}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}}\right)^{\left(\mathrm{1}\right)} =−\mathrm{4}\left(\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{2}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=−\mathrm{4}×\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{4}{x}+\mathrm{2}−\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}{x}+\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }=−\mathrm{4}×\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{8}×\frac{{x}^{\mathrm{2}} −{x}−\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:{so}\:{f}^{'} \left({x}\right)=\mathrm{0}\:\Leftrightarrow{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0}\:{and}\:{x}\neq−\mathrm{1} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{2}\right)=\mathrm{9}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}=\mathrm{2}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}=−\mathrm{1} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:+ \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{incr}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{decr}\:\:\:\:\:{f}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:{incr} \\ $$$${inf}\:{f}\left({x}\right)\:={f}\left(\mathrm{2}\right)=\frac{\mathrm{4}−\mathrm{12}+\mathrm{5}}{\mathrm{4}+\mathrm{4}+\mathrm{1}}\:=\frac{−\mathrm{3}}{\mathrm{9}}\:=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$

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