Find-minimum-value-of-y-2x-x-2-x-1-x-y-R-Without-Differential-

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

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 \dots 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}