Find-range-of-y-x-x-1-x-2-

Question Number 34821 by ajfour last updated on 11/May/18

F i n d r a n g e o f

y = \frac{x}{(x - 1) (x - 2)} .

Commented by ajfour last updated on 11/May/18

Answered by ajfour last updated on 11/May/18

\frac{d y}{d x} = \frac{x^{2} - 3 x + 2 - x (2 x - 3)}{{(x - 1)}^{2} {(x - 2)}^{2}}

\frac{d y}{d x} = 0 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2}

y ∣_{x = - \sqrt{2}} = b = \frac{- \sqrt{2}}{(- \sqrt{2} - 1) (- \sqrt{2} - 2)}

\Rightarrow b = - \frac{1}{{(\sqrt{2} + 1)}^{2}} = - \frac{1}{(3 + 2 \sqrt{2})}

= - (3 - 2 \sqrt{2})

y ∣_{x = \sqrt{2}} = a = \frac{\sqrt{2}}{(\sqrt{2} - 1) (\sqrt{2} - 2)}

\Rightarrow a = - \frac{1}{{(\sqrt{2} - 1)}^{2}}

a = - \frac{1}{3 - 2 \sqrt{2}} = - (3 + 2 \sqrt{2})

S o,

y \in (- \infty, - 3 - 2 \sqrt{2}] \cup [2 \sqrt{2} - 3, \infty) .

Commented by NECx last updated on 12/May/18

w o w \dots . T h i s s t y l e i s n e w t o m e .

T h a n k s

Answered by Joel579 last updated on 11/May/18

y (x^{2} - 3 x + 2) = x

{y x}^{2} - 3 y x - x + 2 y = 0

{y x}^{2} - (3 y + 1) x + 2 y = 0

x = \frac{(3 y + 1) \pm \sqrt{{(3 y + 1)}^{2} - 4 y (2 y)}}{2 y}

= \frac{(3 y + 1) \pm \sqrt{y^{2} + 6 y + 1}}{2 y}

y \neq 0

y^{2} + 6 y + 1 ⩾ 0

y ⩽ - 3 - 2 \sqrt{2} \lor y ⩾ - 3 + 2 \sqrt{2}

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