If x ∈ R, the least value of the expression ((x^2 −6x+5)/(x^2 +2x+1)) is


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Question Number 81672 by zainal tanjung last updated on 14/Feb/20

$If x \in R, the least value of the$ $expression \frac{x^{2} - 6 x + 5}{x^{2} + 2 x + 1} is$
Commented by Kunal12588 last updated on 14/Feb/20

$y = \frac{(x - 5) (x - 1)}{{(x + 1)}^{2}}$ $\frac{d y}{d x} = \frac{{(x + 1)}^{2} [(x - 5) + (x - 1)] - (x - 5) (x - 1) [2 (x + 1)]}{{(x + 1)}^{4}}$ $\Rightarrow \frac{d y}{d x} = \frac{2 (x^{2} + 2 x + 1) (x - 3) - 2 (x - 5) (x^{2} - 1)}{{(x + 1)}^{4}}$ $\Rightarrow \frac{d y}{d x} = \frac{2 [x^{3} + 2 x^{2} + x - 3 x^{2} - 6 x - 3 - (x^{3} - 5 x^{2} - x + 5)]}{{(x + 1)}^{4}}$ $\Rightarrow \frac{d y}{d x} = \frac{2 (4 x^{2} - 4 x - 8)}{{(x + 1)}^{4}}$ $\Rightarrow \frac{d y}{d x} = \frac{8 (x^{2} - x - 2)}{{(x + 1)}^{4}}$ $\Rightarrow \frac{d y}{d x} = \frac{8 (x - 2) (x + 1)}{{(x + 1)}^{4}}$ $f o r m a x o r m i n$ $\frac{d y}{d x} = 0 \Rightarrow x = 2, x \neq - 1$ $\frac{d^{2} y}{{d x}^{2}} > 0 w h e n x = 2$ $y_{l o c a l m i n} = \frac{(2 - 5) (2 - 1)}{{(2 + 1)}^{2}} = \frac{- 3}{3^{2}} = - \frac{1}{3} = - 0 . \overset{―}{3}$
Commented by Tony Lin last updated on 14/Feb/20

$l e t \frac{x^{2} - 6 x + 5}{x^{2} + 2 x + 1} = k$ $\Rightarrow x^{2} - 6 x + 5 = {k x}^{2} + 2 k x + k$ $\Rightarrow (k - 1) x^{2} + (2 k + 6) x + (k - 5) = 0$ $Δ = {(2 k + 6)}^{2} - 4 (k - 1) (k - 5) = 0$ $\Rightarrow k = - \frac{1}{3} w h e n x = 2$ $\frac{x^{2} - 6 x + 5}{x^{2} + 2 x + 1} = - \frac{1}{3} m i n$
Commented by mathmax by abdo last updated on 14/Feb/20

$f (x) = \frac{x^{2} + 2 x + 1 - 8 x + 4}{x^{2} + 2 x + 1} = 1 + \frac{4 - 8 x}{x^{2} + 2 x + 1} \Rightarrow$ $f^{^{'}} (x) = - 4 {(\frac{2 x - 1}{x^{2} + 2 x + 1})}^{(1)} = - 4 (\frac{2 (x^{2} + 2 x + 1) - (2 x - 1) (2 x + 2)}{{(x^{2} + 2 x + 1)}^{2}})$ $= - 4 \times \frac{2 x^{2} + 4 x + 2 - 4 x^{2} - 4 x + 2 x + 2}{{(x + 1)}^{4}} = - 4 \times \frac{- 2 x^{2} + 2 x + 4}{{(x + 1)}^{4}}$ $= 8 \times \frac{x^{2} - x - 2}{{(x + 1)}^{4}} s o f^{^{'}} (x) = 0 \Leftrightarrow x^{2} - x - 2 = 0 a n d x \neq - 1$ $Δ = 1 - 4 (- 2) = 9 \Rightarrow x_{1} = \frac{1 + 3}{2} = 2 a n d x_{2} = \frac{1 - 3}{2} = - 1$ $x - \infty - 1 2 + \infty$ $f^{^{'}} + ∣∣ - 0 +$ $f i n c r d e c r f (2) i n c r$ $i n f f (x) = f (2) = \frac{4 - 12 + 5}{4 + 4 + 1} = \frac{- 3}{9} = - \frac{1}{3}$
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