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Find all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\le11$$ is true.

I know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection of those $3$ cases. The solution will be $-2\ge x\le9$.

But suppose if I interpret this in this way:

What number $x$ satisfy the condition that the distance between $x$ and $6$ plus the distance between $x$ and $1$ is less than or equal to $11$?

I would be better to get an idea to solve these types of problems by geometrically by intuition using number line.

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    $\begingroup$ The geometric interpretation is a very good way of viewing the problem. $\endgroup$ – André Nicolas Mar 12 '16 at 6:41
  • $\begingroup$ As $x$ moves between $(1,6)$, the sum of distances cannot change. Moving away from the interval the sum increases twice as fast as distance to both points increases. That's usually enough to solve this. $\endgroup$ – Macavity Mar 12 '16 at 6:50
  • $\begingroup$ @Macavity I'm usable to visualize how will the sum change twice as fast when $x$ moves away from the interval $(1,6)$ ? $\endgroup$ – Heisenberg Mar 12 '16 at 7:05
  • $\begingroup$ Outside the interval, $x$ is moving away from both points, so the distances naturally add. The interval has measure $5$ and the slack is $11-5=6$, so $x$ can move at best $3$ on either side of the interval. $\endgroup$ – Macavity Mar 12 '16 at 7:08
  • $\begingroup$ One of the inequalities is pointing the wrong way where you write "The solution will be $-2\ge x\le9$." $\endgroup$ – Barry Cipra Mar 12 '16 at 14:38
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As you said, we are looking for points for which:

$$(\text{distance to $1$})+(\text{distance to $6$})\leq 11.$$ The first thing which now comes to my mind is an ellipse.

We first consider all $x\in \mathbb C$ for which $\vert x-1\vert +\vert x-6\vert \leq 11$. All points of this "filled" ellipse which lie on the real axis are the ones we want.

It is not very hard to imagine what these point will be. If we are on the real line then our furthest left point, $x_\ell$, will lie left of $1$. So the distance to $6$ will be at least five. So we have $$\vert x_\ell-1\vert+\vert x_\ell -6\vert=2\vert x_\ell-1\vert+5=11. $$ And since $x_\ell $ is to the left of $1$ this means that $x_\ell=-2$. Analogously, we find that $x_r=9$.

So we have found that $-2\leq x\leq 9$, through geometrical methods.

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$$|1-x| + |x-6| \le 11$$

Let $A, X, B \in \mathbb R^n$ (Euclidean $n$-space). If $X\in \overline{AB}$ ( $X$ is on the line segment $\overline{AB}$ ), we say that $X$ is between $A$ and $B$, and we write this as $A-X-B$.

$$A-X-B \; \text{ if and only if } \; \|A-X\| + \|X-B\| = \|A-B\|$$

On $\mathbb R^1$ (the real number line), every point is on the line through points $1$ and $6$. So there are three possibilities for $1, x$, and $6$ : $x-1-6$, $1-x-6$, or $1-6-x$.

CASE: $x-1-6$

Then $ x \le 1 \le 6$

\begin{align} |1-x| + |x-6| &\le 11 \\ (1-x) + (6-x) &\le 11\\ -2x &\le 4\\ x &\ge -2\\ x &\in [-2,1] \end{align}

CASE: $1-x-6$

Then $1 \le x \le 6$ and $|1-x| + |x-6| = |1-6| = 5$

\begin{align} |1-x| + |x-6| &\le 11 \\ 5 &\le 11 \\ x &\in [1,6] \end{align}

CASE: $1-6-x$

Then $1 \le 6 \le x$

\begin{align} |1-x| + |x-6| &\le 11 \\ (x-1) + (x-6) &\le 11 \\ 2x &\le 18\\ x &\le 9\\ x &\in [6,9] \end{align}

So $x \in [-2,9]$

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  • $\begingroup$ A nit-picker would say: "You did not formally consider the cases $x=1$ and $x=6$." $\endgroup$ – gebruiker Mar 12 '16 at 10:51
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Add graphics solution:

enter image description here

$$y=|x-1|+|x-6|; y=11$$ $$x \in [x_1;x_2] =[-2;9]$$

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