Which one of the given figures P, Q, R and S represents the graph of the following function?
\[
f(x)=\left|\,|x+2|-|x-1|\,\right|
\]
Show Hint
For graphs with nested absolute values, split the real line at all sign-change points, simplify in each interval, and sketch from the resulting linear pieces. Zeros often occur at points equidistant from the inner absolute-value centers.
Step 1: Identify breakpoints of the absolute values.
Breakpoints occur where the inner expressions change sign: $x=-2$ and $x=1$.
Also, $f(x)=0$ when $|x+2|=|x-1|$, i.e., when the point $x$ is equidistant from $-2$ and $1$.
Midpoint $\Rightarrow x=\dfrac{-2+1}{2}=-\dfrac{1}{2}=-0.5$.
Hence $f(-0.5)=0$.
Step 2: Compute $f(x)$ piecewise.
\[
f(x)=\Big||x+2|-|x-1|\Big|=
\begin{cases}
|(x+2)-(x-1)|=|3|=3, & x\ge 1, [4pt]
|(x+2)-(1-x)|=|2x+1|, & -2\le x\le 1, [4pt]
|(-(x+2))-(1-x)|=|-3|=3, & x\le -2.
\end{cases}
\]
Step 3: Shape from the piecewise form.
- For $x\le -2$: $f(x)=3$ (horizontal line).
- For $-2\le x\le 1$: $f(x)=|2x+1|$ — a V-shape with vertex at $x=-0.5$, value $0$, and value $3$ at both $x=-2$ and $x=1$.
- For $x\ge 1$: $f(x)=3$ (horizontal line).
Thus the graph is a flat segment at $y=3$ for $x\le -2$, then a V-shaped dip to $0$ at $x=-0.5$, then back up to $3$ at $x=1$, followed by another flat segment at $y=3$ for $x\ge 1$.
Step 4: Match with the options.
Only figure P shows a valley touching $0$ near $x=-0.5$ and being {constant} at $y=3$ for both $x\le -2$ and $x\ge 1$.
Final Answer:
\[
\boxed{\text{P}}
\]
