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Exercise:

Determine the intervals in which the following inequality is satisfied: $$|1-x|-x\geq 0$$

Attempt:

What to Expect:

A quick manipulation renders the following: $|1-x|\geq x$.

Graphing both sides:

Eyeballing, the answer seems to be: $x \leq \frac{1}{2}$.

Solution:

(1) $|1-x|-x\geq 0$

(2) $|1-x| \geq x$

(3)

  • when $x \geq 0$:
    • when $1-x \geq 0$: $1-x \geq x$
    • when $1-x < 0$: $-(1-x) \geq x$
  • when $x < 0$: $x-1 > 0$

(4)

  • when $x \geq 0$:
    • when $1 \geq x$: $\frac{1}{2} \geq x$
    • when $1 < x$: invalid
  • when $x < 0$: $x > 1$

Request:

I do see the expected answer in (4), but according to my solution it's only applicable when $x \geq 0$. When $x < 0$, I get an answer that seems to have no resemblance in the expected answer. Where and what did I do wrong?

  • 1
    If $x < 0$ you have $x < 0 \le |1-x|$ which is true. – 2017-01-03
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    You only need to consider two cases: (1) when $x \geq 1$ and (2) when $x < 1$. – 2017-01-03
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    @AnuragA -- Yes, I see (as done in velutluna's answer). However, why shouldn't you check when $x\geq 0$ and $x < 0$, because inequalities might be flipped then? – 2017-01-03
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    When $x < 0$ then $|anything| \ge 0 > x$ so always valid – 2017-01-03
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    ..or, if you must. If $x < 0$ then $1 -x = 1 + (-x) > 0$ so $|1 -x| = 1-x \ge x$ so $1 \ge 2x$ so $1 \ge x$. So $x < 0$ and $x \le 1/2$ so .... $x < 1$. – 2017-01-03

3 Answers 3

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When $x \ge 1$, $$|1-x|=x-1\ge x$$ and there is no solution.

When $x \le 1$, $$|1-x|=1-x\ge x$$ $$1\ge 2x$$ $$1/2\ge x$$

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    Why were you able to disregard the RHS $x$, as in: if $x$ is negative then the inequality must be flipped? – 2017-01-03
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    Irrespective of the sign of $x$, $\ge x$ is always $\ge x$. – 2017-01-03
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There are three cases:

$x=1$

the inequation becomes $-1\geq 0$ which is false.

$x>1$

it becomes

$x-1-x=-1\geq 0$ . it is false

$x<1$

it gives $1-x-x\geq 0$ and $x\leq \frac{1}{2}$

the inequation is satisfied in $(-\infty,\frac{1}{2}]$.

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$$|1-x|-x\geq 0$$

$$\Leftrightarrow |1-x|\geq x $$

$$ \Leftrightarrow [(1-x)\geq x] \text{ or } [(1-x)\le -x]$$

$$ \Leftrightarrow (1\geq 2x) \text{ or } (1 \le 0)$$

$$ \Leftrightarrow \frac{1}{2}\geq x \text{ or } 1 \le 0$$

$$ \Leftrightarrow x \le \frac{1}{2} \text{ or } 1 \le 0$$

$$ \Leftrightarrow x \in ]-\infty, \frac{1}{2}] \text{ or } x \in \emptyset $$

$$ \Leftrightarrow x \in ]-\infty, \frac{1}{2}] \cup \emptyset $$

$$ \Leftrightarrow x \in ]-\infty, \frac{1}{2}]$$