how to find all values of x for which the tangent to y=x-(1/x) is parallel to the line 2x-y=5?

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Answer 1 · 2018-06-01T10:02:24

$x = \pm 1$ $x=+-1$

Given curve is-
$y = x - \frac{1}{x}$ $y= x-1/x$

$\frac{d y}{d x} = 1 + \frac{1}{x^{2}}$ $dy/dx= 1+1/x^2$

$S l o p e = 1 + \frac{1}{x^{2}}$ $Slope= 1+1/x^2$

Slope of the line $2 x - y = 5$ $2x-y=5$ is $2$ $2$

By giving condition,

$2 = 1 + \frac{1}{x^{2}}$ $2= 1+1/x^2$

$1 = \frac{1}{x^{2}}$ $1= 1/x^2$

$1 = x^{2}$ $1=x^2$

$x = \pm 1$ $x=+-1$