What is the minimum value of |x + 1| + |x − 2|?

Step-by-step solution

1. 1

   Read each absolute value as a distance. Since $|x-a|$ is the distance from $x$ to $a$ on the number line, rewrite the expression as

   $|x-(-1)|+|x-2|$

   so it is the total distance from the point $x$ to the two fixed points $-1$ and $2$. This reframing replaces a three-case algebraic split with one geometric fact.

2. 2

   Reason about where the total distance is smallest. If $x$ lies between $-1$ and $2$, the two distances add up to exactly the gap between the points. If $x$ lies outside that interval, you must first travel back, so the total is strictly larger.

3. 3

   Compute the gap.

   $2-(-1)=3$

   so the minimum value is $3$, attained for every $x$ in $[-1,2]$ — not at a single point.

   $\boxed{3}$

4. 4

   Confirm with the algebraic case split. For $x<-1$: $-(x+1)-(x-2)=1-2x>3$. For $-1\le x\le 2$: $(x+1)-(x-2)=3$. For $x>2$: $(x+1)+(x-2)=2x-1>3$. The three pieces confirm the value $3$ and the flat middle section.

5. 5

   Cross-check with the triangle inequality. $|x+1|+|x-2|=|x+1|+|2-x|\ge|(x+1)+(2-x)|=3$, with equality exactly when $x+1$ and $2-x$ have the same sign — i.e. when $-1\le x\le 2$, matching the case split. Spot values: at $x=0$ the sum is $1+2=3$; at $x=5$ it is $6+3=9$.