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Question Number 141294 by bemath last updated on 17/May/21
Find max & min value of f(x)=(x/(x^2 −5x+9)).
$${Find}\:{max}\:\&\:{min}\:{value}\:{of} \\ $$$$\:{f}\left({x}\right)=\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{9}}. \\ $$
Answered by MJS_new last updated on 17/May/21
(df/dx)=((−x^2 +9)/((x^2 −5x+9)^2 ))=0 ⇒ x=±3 min at x=−3 max at x=3
$$\frac{{df}}{{dx}}=\frac{−{x}^{\mathrm{2}} +\mathrm{9}}{\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{9}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{3} \\ $$$$\mathrm{min}\:\mathrm{at}\:{x}=−\mathrm{3} \\ $$$$\mathrm{max}\:\mathrm{at}\:{x}=\mathrm{3} \\ $$
Answered by EDWIN88 last updated on 17/May/21
Domain f(x): ∀xR ((df(x))/dx) = ((x^2 −5x+9−x(2x−5))/((x^2 −5x+9)^2 )) = 0 ((df(x))/dx) = ((−x^2 +9)/((x^2 −5x+9)^2 )) ⇒−x^2 +9 = 0 , x=± 3 by first−derivative tes we get decreasing in x<−3 ∨ x>3 increasing in −3<x<3 thus { ((min at x=−3)),((max at x=3)) :}⇒ { ((f_(min) = −(3/(33))=−(1/(11)))),((f_(max) = (3/3)=1)) :}
$$\:\mathrm{Domain}\:\mathrm{f}\left(\mathrm{x}\right):\:\forall\mathrm{x}\mathbb{R} \\ $$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{9}−\mathrm{x}\left(\mathrm{2x}−\mathrm{5}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{9}\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\frac{−\mathrm{x}^{\mathrm{2}} +\mathrm{9}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{9}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow−\mathrm{x}^{\mathrm{2}} +\mathrm{9}\:=\:\mathrm{0}\:,\:\mathrm{x}=\pm\:\mathrm{3} \\ $$$$\mathrm{by}\:\mathrm{first}−\mathrm{derivative}\:\mathrm{tes}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\mathrm{decreasing}\:\mathrm{in}\:\mathrm{x}<−\mathrm{3}\:\vee\:\mathrm{x}>\mathrm{3} \\ $$$$\mathrm{increasing}\:\mathrm{in}\:−\mathrm{3}<\mathrm{x}<\mathrm{3} \\ $$$$\mathrm{thus}\:\begin{cases}{\mathrm{min}\:\mathrm{at}\:\mathrm{x}=−\mathrm{3}}\\{\mathrm{max}\:\mathrm{at}\:\mathrm{x}=\mathrm{3}}\end{cases}\Rightarrow\begin{cases}{\mathrm{f}_{\mathrm{min}} \:=\:−\frac{\mathrm{3}}{\mathrm{33}}=−\frac{\mathrm{1}}{\mathrm{11}}}\\{\mathrm{f}_{\mathrm{max}} \:=\:\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{1}}\end{cases} \\ $$
Answered by bemath last updated on 17/May/21
y = (x/(x^2 −5x+9)) ⇒yx^2 −5xy+9y−x =0 ⇒yx^2 −(5y+1)x+9y =0 Δ=(5y+1)^2 −4(y)(9y)≥0 ⇒25y^2 +10y+1−36y^2 ≥ 0 ⇒−11y^2 +10y+1 ≥ 0 ⇒11y^2 −10y−1≤0 ⇒(11y+1)(y−1)≤0 ⇒−(1/(11)) ≤ y ≤ 1 → { ((min =−(1/(11)))),((max = 1)) :}
$$\:{y}\:=\:\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{9}} \\ $$$$\Rightarrow{yx}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{9}{y}−{x}\:=\mathrm{0} \\ $$$$\Rightarrow{yx}^{\mathrm{2}} −\left(\mathrm{5}{y}+\mathrm{1}\right){x}+\mathrm{9}{y}\:=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{5}{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({y}\right)\left(\mathrm{9}{y}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{25}{y}^{\mathrm{2}} +\mathrm{10}{y}+\mathrm{1}−\mathrm{36}{y}^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow−\mathrm{11}{y}^{\mathrm{2}} +\mathrm{10}{y}+\mathrm{1}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{11}{y}^{\mathrm{2}} −\mathrm{10}{y}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{11}{y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{11}}\:\leqslant\:{y}\:\leqslant\:\mathrm{1} \\ $$$$\:\rightarrow\begin{cases}{{min}\:=−\frac{\mathrm{1}}{\mathrm{11}}}\\{{max}\:=\:\mathrm{1}}\end{cases} \\ $$
Answered by mr W last updated on 17/May/21
f(x)=(1/(x+(9/x)−5)) x+(9/x)≥2(√9)=6 ⇒f(x)_(max) =(1/(6−5))=1 when x=3 x+(9/x)≤−2(√9)=−6 ⇒f(x)_(min) =(1/(−6−5))=−(1/(11)) when x=−3
$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\frac{\mathrm{9}}{{x}}−\mathrm{5}} \\ $$$${x}+\frac{\mathrm{9}}{{x}}\geqslant\mathrm{2}\sqrt{\mathrm{9}}=\mathrm{6} \\ $$$$\Rightarrow{f}\left({x}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{6}−\mathrm{5}}=\mathrm{1}\:{when}\:{x}=\mathrm{3} \\ $$$${x}+\frac{\mathrm{9}}{{x}}\leqslant−\mathrm{2}\sqrt{\mathrm{9}}=−\mathrm{6} \\ $$$$\Rightarrow{f}\left({x}\right)_{{min}} =\frac{\mathrm{1}}{−\mathrm{6}−\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{11}}\:{when}\:{x}=−\mathrm{3} \\ $$
Answered by 1549442205PVT last updated on 06/Jun/21
put A=(x/(x^2 −5x+9))⇔Ax^2 −(5A+1)x+9A=0 This quadratc equation always has roots so Δ=(5A+1)^2 −36A^2 ≥0 −11A^2 +10A+1≥0⇔(1+11A)(1−A)≥0 ⇔((−1)/(11))≤A≤1.Hence,A_(min) =((−1)/(11)) when x=−3 A_(max) =2 when x=3
$$\mathrm{put}\:\mathrm{A}=\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{9}}\Leftrightarrow\mathrm{Ax}^{\mathrm{2}} −\left(\mathrm{5A}+\mathrm{1}\right)\mathrm{x}+\mathrm{9A}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{quadratc}\:\:\mathrm{equation}\:\mathrm{always}\:\mathrm{has} \\ $$$$\mathrm{roots}\:\mathrm{so}\:\Delta=\left(\mathrm{5A}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{36A}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$−\mathrm{11A}^{\mathrm{2}} +\mathrm{10A}+\mathrm{1}\geqslant\mathrm{0}\Leftrightarrow\left(\mathrm{1}+\mathrm{11A}\right)\left(\mathrm{1}−\mathrm{A}\right)\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\frac{−\mathrm{1}}{\mathrm{11}}\leqslant\mathrm{A}\leqslant\mathrm{1}.\mathrm{Hence},\mathrm{A}_{\mathrm{min}} =\frac{−\mathrm{1}}{\mathrm{11}}\:\mathrm{when}\:\mathrm{x}=−\mathrm{3} \\ $$$$\mathrm{A}_{\mathrm{max}} =\mathrm{2}\:\mathrm{when}\:\mathrm{x}=\mathrm{3} \\ $$$$ \\ $$

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