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

Sep 21, 2016

$y = x$

Explanation:

For $x > 0$, we have $\left\mid x \right\mid = x$, so $f \left(x\right) = \frac{1}{x}$ and $y = 1$.

Furthermore, $f ' \left(x\right) = - \frac{1}{x} ^ 2$.

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

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

$y = x$.

Sep 21, 2016

$y = x$

Explanation:

See image at end of post....

we can see that at $x = 1$ we have $\frac{1}{\left\mid x \right\mid} = \frac{1}{x}$

and so the slope is $\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right) {\setminus}_{x = 1} = {\left(- \frac{1}{x} ^ 2\right)}_{x = 1} = - 1$

Thus the tangent vector is $\vec{T} = \left(\begin{matrix}1 \\ - 1\end{matrix}\right)$

Therefore, for the normal vector $\vec{N} = \left(\begin{matrix}a \\ b\end{matrix}\right)$ we have

$\left(\begin{matrix}- 1 \\ 1\end{matrix}\right) \cdot \left(\begin{matrix}a \\ b\end{matrix}\right) = 0 \implies a = b$ so we can choose $a = b = 1$ so that $\vec{N} = \left(\begin{matrix}1 \\ 1\end{matrix}\right)$

the slope of the normal is therefore $\frac{1}{1} = 1$

we can also say that $y \left(1\right) = 1$

so we have

$m = \frac{y - {y}_{o}}{x - {x}_{o}}$

$1 = \frac{y - 1}{x - 1}$

or $y = x$