What is the equation of the line normal to $f(x)=1/abs(x)$ at $x=1$ ?

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Answer 1 · 2016-09-21T19:36:06

$y=x$

For $x > 0$ , we have $absx = x$ , so $f(x) = 1/x$ and $y=1$ .

Furthermore, $f'(x) = -1/x^2$ .

At $x=1$ the slope of the tangent line is $f'(1) = -1$ the slope of the normal line is the opposite reciprocal, so $m_(norm) = -1/((-1)) = 1$ .

The equation of the line through $(1,1)$ with slope $1$ is

$y=x$ .

Answer 2 · 2016-09-21T19:38:07

$y = x$

See image at end of post....

we can see that at $x = 1$ we have $1/absx = 1/x$

and so the slope is $d/dx (1/x)_(x = 1) = (- 1/x^2)_(x = 1) = -1$

Thus the tangent vector is $vec T = ((1), (-1))$

Therefore, for the normal vector $vec N = ((a),(b))$ we have

$((-1),(1)) * ((a),(b)) = 0 implies a = b$ so we can choose $a = b = 1$ so that $vec N = ((1),(1))$

the slope of the normal is therefore $1/1 = 1$

we can also say that $y(1) = 1$

so we have

$m = (y - y_o)/(x - x_o)$

$1 = (y - 1)/(x - 1)$

or $y = x$

enter image source here

What is the equation of the line normal to f(x)=1/abs(x) f(x)=1/abs(x) at x=1x=1?