Find minimum value of y = ((2x)/(x^2 + x + 1)) x , y ∈ R Without Differential


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Question Number 78628 by naka3546 last updated on 19/Jan/20

$F i n d m i n i m u m v a l u e o f$ $y = \frac{2 x}{x^{2} + x + 1}$ $x, y \in R$ $W i t h o u t D i f f e r e n t i a l$
Commented by john santu last updated on 19/Jan/20
$denominator x^2 +x+1 >0 ∀x∈R x^2 y+xy+y−2x=0 x^2 y+(y−2)x+y=0 △= (y−2)^2 −4y.y≥0 (y−2)^2 −(2y)^2 ≥0 (3y−2)(−y−2)≥0 (3y−2)(y+2)≤0 −2≤y≤(2/3) { ((minimum = −2)),((maximum = (2/3))) :}$
$d e n o m i n a t o r x^{2} + x + 1 > 0 \forall x \in R$ $x^{2} y + x y + y - 2 x = 0$ $x^{2} y + (y - 2) x + y = 0$ $△ = {(y - 2)}^{2} - 4 y . y ⩾ 0$ ${(y - 2)}^{2} - {(2 y)}^{2} ⩾ 0$ $(3 y - 2) (- y - 2) ⩾ 0$ $(3 y - 2) (y + 2) ⩽ 0$ $- 2 ⩽ y ⩽ \frac{2}{3} {\begin{cases} m i n i m u m = - 2 \\ m a x i m u m = \frac{2}{3} \end{cases}$
	Answered by jagoll last updated on 19/Jan/20

	$\lim_{x \to \infty} \frac{2 x}{x^{2} + x + 1} = 0$ $\lim_{x \to - \infty} \frac{2 x}{x^{2} + x + 1} = 0$ $y_{\min} = 0$
	Commented by mr W last updated on 19/Jan/20

	$w h a t d o y o u g e t w h e n y o u p u t x = - 1 ?$
	Commented by jagoll last updated on 19/Jan/20

	$x = - 1 \Rightarrow y = \frac{- 2}{1} = - 2 sir ?$
	Commented by jagoll last updated on 19/Jan/20

	$well . . . what minimum value sir ?$
	Commented by mr W last updated on 19/Jan/20

	$i f x > 0 : y = \frac{2}{1 + x + \frac{1}{x}} = \frac{2}{3 + {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2}}$ $i f x < 0 : y = \frac{2}{1 + x + \frac{1}{x}} = \frac{2}{- 1 - {(\sqrt{- x} - \frac{1}{\sqrt{- x}})}^{2}}$ $y (0) = 0 a n d \lim_{x \to + \infty} y = 0, y (x > 0) > 0$ $\Rightarrow t h e r e i s a m a x i m u m f o r x > 0$ $i t i s t o s e e w h e n \sqrt{x} = \frac{1}{\sqrt{x}}, i . e . x = 1, y_{m a x} = \frac{2}{3}$ $y (0) = 0 a n d \lim_{x \to - \infty} y = 0, y (x < 0) < 0$ $\Rightarrow t h e r e i s a m i n i m u m f o r x < 0$ $i t i s t o s e e w h e n \sqrt{- x} = \frac{1}{\sqrt{- x}}, i . e . x = - 1, y_{m i n} = \frac{2}{- 1} = - 2$
	Commented by john santu last updated on 19/Jan/20

	$t h e l a s t l i n e y_{m i n} o r y_{m a x} ?$
	Commented by mr W last updated on 19/Jan/20

	$t y p o . y_{m i n} s h o u l d b e .$
	Answered by behi83417@gmail.com last updated on 19/Jan/20

	$\frac{1}{y} = \frac{x^{2} + x + 1}{2 x} = \frac{1}{2} (x + \frac{1}{x} + 1) ⩾ \frac{1}{2} (2 + 1) = \frac{3}{2}$ $\Rightarrow y ⩽ \frac{2}{3}$
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