Question

If \( |x-1| + |x-3| \leq 8 \), then the values of \( x \) lie in the interval:

• A. $(-\infty, -2]$
• B. $[-2, 6]$
• C. $(-3, 7)$
• D. $(-2, \infty)$
• E. $[6, \infty)$

Hint:

Geometrically, $|x-a| + |x-b| = k$ represents the sum of distances from $x$ to $a$ and $b$. If $k$ is greater than the distance between $a$ and $b$, the solution will always be a closed interval centered at the midpoint of $a$ and $b$.

Answer 1

The Correct Option is B

Concept: Absolute value inequalities can be solved by considering different cases based on the critical points where the expressions inside the absolute values become zero. Here, the critical points are $x=1$ and $x=3$.

Step 1: Case 1: $x < 1$.
Both expressions are negative: $-(x-1) - (x-3) \leq 8$ \[ -x + 1 - x + 3 \leq 8 \Rightarrow -2x + 4 \leq 8 \] \[ -2x \leq 4 \Rightarrow x \geq -2 \] So, $-2 \leq x < 1$.

Step 2: Case 2: $1 \leq x < 3$.
The first is positive, second is negative: $(x-1) - (x-3) \leq 8$ \[ x - 1 - x + 3 \leq 8 \Rightarrow 2 \leq 8 \] This is always true for the entire interval $[1, 3)$.

Step 3: Case 3: $x \geq 3$.
Both are positive: $(x-1) + (x-3) \leq 8$ \[ 2x - 4 \leq 8 \Rightarrow 2x \leq 12 \Rightarrow x \leq 6 \] So, $3 \leq x \leq 6$.

Step 4: Combine the intervals.
The union of $[-2, 1)$, $[1, 3)$, and $[3, 6]$ results in the interval $[-2, 6]$.