Question

Which one of the given figures P, Q, R and S represents the graph of the following function? 
\( f(x) = | |x + 2| - |x - 1| | \) 

 

• A. P
• B. Q
• C. R
  (D) S
• D.

Hint:

When dealing with functions involving \(|x-a|\) and \(|x-b|\), always use the points \(x=a\) and \(x=b\) as critical points to define the intervals for your piecewise analysis. This simplifies the problem significantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires plotting a function that involves nested absolute values. The key is to analyze the function piecewise by considering different intervals based on the points where the expressions inside the absolute value signs become zero.
Step 2: Key Formula or Approach:
The definition of absolute value is \(|a| = a\) if \(a \ge 0\) and \(|a| = -a\) if \(a<0\).
The critical points for the expressions inside the inner absolute values are where \(x+2=0\) and \(x-1=0\). These points are \(x = -2\) and \(x = 1\). These points divide the number line into three intervals: \(x<-2\), \(-2 \le x<1\), and \(x \ge 1\). We will analyze the function \(f(x)\) in each interval.
Step 3: Detailed Calculation:
Case 1: \(x<-2\)
In this interval, \(x+2<0\) and \(x-1<0\).
So, \(|x+2| = -(x+2)\) and \(|x-1| = -(x-1)\).
\[ f(x) = | (-(x+2)) - (-(x-1)) | \] \[ f(x) = | -x - 2 + x - 1 | \] \[ f(x) = | -3 | = 3 \] So, for \(x<-2\), the graph is a horizontal line at \(y=3\).
Case 2: \(-2 \le x<1\)
In this interval, \(x+2 \ge 0\) and \(x-1<0\).
So, \(|x+2| = x+2\) and \(|x-1| = -(x-1)\).
\[ f(x) = | (x+2) - (-(x-1)) | \] \[ f(x) = | x + 2 + x - 1 | \] \[ f(x) = | 2x + 1 | \] This is a V-shaped graph with its vertex at \(2x+1=0\), which is \(x = -1/2\). At this point, \(f(-1/2) = 0\).
At the endpoints of the interval:
\(f(-2) = |2(-2)+1| = |-3| = 3\). \(f(1)\) (approaching from the left) would be \(|2(1)+1| = |3| = 3\).
Case 3: \(x \ge 1\)
In this interval, \(x+2>0\) and \(x-1 \ge 0\).
So, \(|x+2| = x+2\) and \(|x-1| = x-1\).
\[ f(x) = | (x+2) - (x-1) | \] \[ f(x) = | x + 2 - x + 1 | \] \[ f(x) = | 3 | = 3 \] So, for \(x \ge 1\), the graph is a horizontal line at \(y=3\).
Summary of the graph's shape:
- For \(x<-2\), it's a horizontal line \(y=3\).
- From \(x=-2\) to \(x=1\), it's a V-shape \(y=|2x+1|\) that goes from \(y=3\) down to \(y=0\) (at \(x=-1/2\)) and back up to \(y=3\).
- For \(x \ge 1\), it's a horizontal line \(y=3\).
This description perfectly matches the graph in figure P.
Step 4: Final Answer:
The function is represented by the graph P.
Step 5: Why This is Correct:
The piecewise analysis correctly breaks down the complex absolute value function. The resulting shape—constant at 3, dipping to 0 in a V-shape between -2 and 1, and then constant at 3 again—is exactly what is shown in graph P.