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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)).
Findmax&minvalueoff(x)=xx2−5x+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
dfdx=−x2+9(x2−5x+9)2=0⇒x=±3minatx=−3maxatx=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)) :}
Domainf(x):∀xRdf(x)dx=x2−5x+9−x(2x−5)(x2−5x+9)2=0df(x)dx=−x2+9(x2−5x+9)2⇒−x2+9=0,x=±3byfirst−derivativeteswegetdecreasinginx<−3∨x>3increasingin−3<x<3thus{minatx=−3maxatx=3⇒{fmin=−333=−111fmax=33=1
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=xx2−5x+9⇒yx2−5xy+9y−x=0⇒yx2−(5y+1)x+9y=0Δ=(5y+1)2−4(y)(9y)⩾0⇒25y2+10y+1−36y2⩾0⇒−11y2+10y+1⩾0⇒11y2−10y−1⩽0⇒(11y+1)(y−1)⩽0⇒−111⩽y⩽1→{min=−111max=1
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(x)=1x+9x−5x+9x⩾29=6⇒f(x)max=16−5=1whenx=3x+9x⩽−29=−6⇒f(x)min=1−6−5=−111whenx=−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
putA=xx2−5x+9⇔Ax2−(5A+1)x+9A=0ThisquadratcequationalwayshasrootssoΔ=(5A+1)2−36A2⩾0−11A2+10A+1⩾0⇔(1+11A)(1−A)⩾0⇔−111⩽A⩽1.Hence,Amin=−111whenx=−3Amax=2whenx=3

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