Question

Which one of the given figures P, Q, R, and S represents the graph of the following function? \[ f(x) = |\,|x+2| - |x-1|\,| \] \begin{center} \includegraphics[width=0.5\textwidth]{02.jpeg} \end{center}

• A. P
• B. Q
• C. R
• D. S

Hint:

When solving absolute value graph problems, always split into regions around critical points and simplify piecewise.

Answer 1

The Correct Option is C

Step 1: Identify critical points. Absolute value expressions change at \(x = -2\) and \(x = 1\). So we break into intervals: \(-\infty < x < -2\), \(-2 \leq x < 1\), and \(x \geq 1\).

Step 2: Case 1: \(x \leq -2\). \(|x+2| = -(x+2), \quad |x-1| = -(x-1)\). So, \[ f(x) = |[-(x+2)] - [-(x-1)]| = | -x-2 + x -1 | = |-3| = 3 \] Thus, \(f(x) = 3\) (constant).

Step 3: Case 2: \(-2 \leq x < 1\). \(|x+2| = x+2, \quad |x-1| = -(x-1)\). So, \[ f(x) = |(x+2) - (-(x-1))| = |x+2 + x -1| = |2x+1| \] For \(-2 \leq x < 1\), \(2x+1\) varies from \(-3\) to \(3\). Thus, \(f(x) = |2x+1|\), V-shaped graph passing through \((-0.5, 0)\).

Step 4: Case 3: \(x \geq 1\). \(|x+2| = x+2, \quad |x-1| = x-1\). So, \[ f(x) = |(x+2) - (x-1)| = |3| = 3 \] Thus, \(f(x) = 3\) (constant).

Step 5: Combine. - For \(x \leq -2\): flat line at 3. - For \(-2 \leq x < 1\): V-shaped \(|2x+1|\). - For \(x \geq 1\): flat line at 3. This matches the shape of **Graph R** in the options.

Final Answer: \[ \boxed{R} \]