# How do you solve and write the following in interval notation: absx> abs(x-1)?

Oct 28, 2017

See below.

#### Explanation:

$\left\mid x \right\mid - {\epsilon}^{2} = \left\mid x - 1 \right\mid$ with $\epsilon \ne 0$ or

$\left\mid x \right\mid - \left\mid x - 1 \right\mid = {\epsilon}^{2}$ or

${\left(\left\mid x \right\mid - \left\mid x - 1 \right\mid\right)}^{2} = {\epsilon}^{4}$ or

${x}^{2} + {\left(x - 1\right)}^{2} - 2 \left\mid x \right\mid \left\mid x - 1 \right\mid = {\epsilon}^{4}$ or

$4 {x}^{2} {\left(x - 1\right)}^{2} = {\left({x}^{2} + {\left(x - 1\right)}^{2} - {\epsilon}^{4}\right)}^{2}$

$4 {x}^{2} {\left(x - 1\right)}^{2} - {\left({x}^{2} + {\left(x - 1\right)}^{2} - {\epsilon}^{4}\right)}^{2} = 0$ or

$\left(1 - {\epsilon}^{4}\right) \left(2 x - 1 - {\epsilon}^{2}\right) \left(2 x - 1 + {\epsilon}^{2}\right) = 0$ or

$\left\{\begin{matrix}2 x - 1 - {\epsilon}^{2} = 0 \\ 2 x - 1 + {\epsilon}^{2} = 0\end{matrix}\right.$ or

$\left\{\begin{matrix}2 x - 1 > 0 \\ 2 x - 1 < 0\end{matrix}\right.$

so finally the inequality $\left\mid x \right\mid > \left\mid x - 1 \right\mid$ has not real solution and consequently the represented set is void.