Let `f(x)=x+1 and phi(x)=x-2.` Then the value of x satisfying `|f(x)+phi(x)|=|f(x)|+|phi(x)|` are :

Options

A. `(-oo,1]`
B. `[2,oo)`
C. `(-oo,-2]`
D. `[1,oo)`

Answer

To solve the problem, we need to find the values of \( x \) that satisfy the equation:

\[ |f(x) + \phi(x)| = |f(x)| + |\phi(x)| \]

where \( f(x) = x + 1 \) and \( \phi(x) = x - 2 \).

### Step 1: Substitute the functions into the equation

First, we substitute \( f(x) \) and \( \phi(x) \) into the equation:

\[ |f(x) + \phi(x)| = |(x + 1) + (x - 2)| = |2x - 1| \]

And for the right-hand side:

\[ |f(x)| + |\phi(x)| = |x + 1| + |x - 2| \]

So we have:

\[ |2x - 1| = |x + 1| + |x - 2| \]

### Step 2: Analyze the absolute values

To solve this equation, we need to consider the critical points where the expressions inside the absolute values change signs. The critical points are:

1. \( 2x - 1 = 0 \) ⇒ \( x = \frac{1}{2} \)
2. \( x + 1 = 0 \) ⇒ \( x = -1 \)
3. \( x - 2 = 0 \) ⇒ \( x = 2 \)

Thus, we will analyze the intervals determined by these points: \( (-\infty, -1) \), \( [-1, \frac{1}{2}) \), \( [\frac{1}{2}, 2) \), and \( [2, \infty) \).

### Step 3: Case 1: \( x < -1 \)

In this interval, all expressions are negative:

\[ |2x - 1| = -(2x - 1) = -2x + 1 \]
\[ |x + 1| = -(x + 1) = -x - 1 \]
\[ |x - 2| = -(x - 2) = -x + 2 \]

Substituting these into the equation gives:

\[ -2x + 1 = (-x - 1) + (-x + 2) \]
\[ -2x + 1 = -2x + 1 \]

This is true for all \( x < -1 \).

### Step 4: Case 2: \( -1 \leq x < \frac{1}{2} \)

In this interval:

\[ |2x - 1| = -(2x - 1) = -2x + 1 \]
\[ |x + 1| = x + 1 \]
\[ |x - 2| = -(x - 2) = -x + 2 \]

Substituting gives:

\[ -2x + 1 = (x + 1) + (-x + 2) \]
\[ -2x + 1 = 3 \]
\[ -2x = 2 \implies x = -1 \]

This value is valid in the interval.

### Step 5: Case 3: \( \frac{1}{2} \leq x < 2 \)

In this interval:

\[ |2x - 1| = 2x - 1 \]
\[ |x + 1| = x + 1 \]
\[ |x - 2| = -(x - 2) = -x + 2 \]

Substituting gives:

\[ 2x - 1 = (x + 1) + (-x + 2) \]
\[ 2x - 1 = 3 \]
\[ 2x = 4 \implies x = 2 \]

This value is not valid in the interval since \( x \) must be less than 2.

### Step 6: Case 4: \( x \geq 2 \)

In this interval:

\[ |2x - 1| = 2x - 1 \]
\[ |x + 1| = x + 1 \]
\[ |x - 2| = x - 2 \]

Substituting gives:

\[ 2x - 1 = (x + 1) + (x - 2) \]
\[ 2x - 1 = 2x - 1 \]

This is true for all \( x \geq 2 \).

### Conclusion

The values of \( x \) satisfying the original equation are:

\[ x \in (-\infty, -1] \cup [2, \infty) \]