Question #53487

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Question #53487

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Answer 1 · 2017-11-21T13:49:05

$f_{(x)} : x^{3} - 4 x^{2} - 1 = 0$ $f_((x)):x^3-4x^2-1=0$

Explanation:

$f_{(x)} : \frac{2 x}{x - 1} - \frac{3 x}{x + 1} = \frac{1}{x}$ $f_((x)):(2x)/(x-1)-(3x)/(x+1)=1/x$

$x \neq 0, \pm 1$ $x!=0,+-1$

$\Rightarrow$ $=>$
$\frac{2 x \cdot (x (x + 1) (x - 1))}{x - 1} - \frac{3 x \cdot (x (x + 1) (x - 1))}{x + 1} = \frac{1 \cdot (x (x + 1) (x - 1))}{x}$ $(2x*(x(x+1)(x-1)))/(x-1)-(3x*(x(x+1)(x-1)))/(x+1)=(1*(x(x+1)(x-1)))/x$

$\Rightarrow 2 x^{2} (x + 1) - 3 x^{2} (x - 1) = x^{2} - 1$ $=> 2x^2(x+1)-3x^2(x-1)=x^2-1$

$\Rightarrow 2 x^{3} + 2 x^{2} - 3 x^{3} + 3 x^{2} = x^{2} - 1$ $=> 2x^3+2x^2-3x^3+3x^2=x^2-1$

$\Rightarrow - x^{3} + 5 x^{2} = x^{2} - 1$ $=> -x^3+5x^2=x^2-1$

$\Rightarrow - x^{3} + 4 x^{2} + 1 = 0$ $=> -x^3+4x^2+1=0$

$\Rightarrow x^{3} - 4 x^{2} - 1 = 0$ $=> x^3-4x^2-1=0$

$\Rightarrow x^{2} (x - 4) = 1$ $=> x^2(x-4)=1$

$(\Rightarrow$ $(=>$ For $x in RR => lim_(x rarr~ 4.0606)(x^3-4x^2-1)~~0)$

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Question #53487

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