Find max & min value of f(x)=(x/(x^2 −5x+9)).


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Question Number 141294 by bemath last updated on 17/May/21

$F i n d m a x & m i n v a l u e o f$ $f (x) = \frac{x}{x^{2} - 5 x + 9} .$
	Answered by MJS_new last updated on 17/May/21

	$\frac{d f}{d x} = \frac{- x^{2} + 9}{{(x^{2} - 5 x + 9)}^{2}} = 0 \Rightarrow x = \pm 3$ $\min at x = - 3$ $\max at x = 3$
	Answered by EDWIN88 last updated on 17/May/21
	$Domain f(x): ∀xR ((df(x))/dx) = ((x^2 −5x+9−x(2x−5))/((x^2 −5x+9)^2 )) = 0 ((df(x))/dx) = ((−x^2 +9)/((x^2 −5x+9)^2 )) ⇒−x^2 +9 = 0 , x=± 3 by first−derivative tes we get decreasing in x<−3 ∨ x>3 increasing in −3<x<3 thus { ((min at x=−3)),((max at x=3)) :}⇒ { ((f_(min) = −(3/(33))=−(1/(11)))),((f_(max) = (3/3)=1)) :}$
	$Domain f (x) : \forall x R$ $\frac{df (x)}{dx} = \frac{x^{2} - 5 x + 9 - x (2 x - 5)}{{(x^{2} - 5 x + 9)}^{2}} = 0$ $\frac{df (x)}{dx} = \frac{- x^{2} + 9}{{(x^{2} - 5 x + 9)}^{2}}$ $\Rightarrow - x^{2} + 9 = 0, x = \pm 3$ $by first - derivative tes we get$ $decreasing in x < - 3 \lor x > 3$ $increasing in - 3 < x < 3$ $thus {\begin{cases} \min at x = - 3 \\ \max at x = 3 \end{cases} \Rightarrow {\begin{cases} f_{\min} = - \frac{3}{33} = - \frac{1}{11} \\ f_{\max} = \frac{3}{3} = 1 \end{cases}$
	Answered by bemath last updated on 17/May/21
	$y = (x/(x^2 −5x+9)) ⇒yx^2 −5xy+9y−x =0 ⇒yx^2 −(5y+1)x+9y =0 Δ=(5y+1)^2 −4(y)(9y)≥0 ⇒25y^2 +10y+1−36y^2 ≥ 0 ⇒−11y^2 +10y+1 ≥ 0 ⇒11y^2 −10y−1≤0 ⇒(11y+1)(y−1)≤0 ⇒−(1/(11)) ≤ y ≤ 1 → { ((min =−(1/(11)))),((max = 1)) :}$
	$y = \frac{x}{x^{2} - 5 x + 9}$ $\Rightarrow {y x}^{2} - 5 x y + 9 y - x = 0$ $\Rightarrow {y x}^{2} - (5 y + 1) x + 9 y = 0$ $Δ = {(5 y + 1)}^{2} - 4 (y) (9 y) ⩾ 0$ $\Rightarrow 25 y^{2} + 10 y + 1 - 36 y^{2} ⩾ 0$ $\Rightarrow - 11 y^{2} + 10 y + 1 ⩾ 0$ $\Rightarrow 11 y^{2} - 10 y - 1 ⩽ 0$ $\Rightarrow (11 y + 1) (y - 1) ⩽ 0$ $\Rightarrow - \frac{1}{11} ⩽ y ⩽ 1$ $\to {\begin{cases} m i n = - \frac{1}{11} \\ m a x = 1 \end{cases}$
	Answered by mr W last updated on 17/May/21

	$f (x) = \frac{1}{x + \frac{9}{x} - 5}$ $x + \frac{9}{x} ⩾ 2 \sqrt{9} = 6$ $\Rightarrow f {(x)}_{m a x} = \frac{1}{6 - 5} = 1 w h e n x = 3$ $x + \frac{9}{x} ⩽ - 2 \sqrt{9} = - 6$ $\Rightarrow f {(x)}_{m i n} = \frac{1}{- 6 - 5} = - \frac{1}{11} w h e n x = - 3$
	Answered by 1549442205PVT last updated on 06/Jun/21

	$put A = \frac{x}{x^{2} - 5 x + 9} \Leftrightarrow {Ax}^{2} - (5 A + 1) x + 9 A = 0$ $This quadratc equation always has$ $roots so Δ = {(5 A + 1)}^{2} - {36 A}^{2} ⩾ 0$ $- {11 A}^{2} + 10 A + 1 ⩾ 0 \Leftrightarrow (1 + 11 A) (1 - A) ⩾ 0$ $\Leftrightarrow \frac{- 1}{11} ⩽ A ⩽ 1 . Hence, A_{\min} = \frac{- 1}{11} when x = - 3$ $A_{\max} = 2 when x = 3$
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