Let f(x)=x amd g(x)=|x| for all `x in R`. Then the function `phi(x)"satisfying"{phi(x)-f(x)}^(2)+{phi(x)-g(x)}^(2)` =0 is

Options

A. `phi (x)=x, x in [0,oo)`
B. `phi (x)=x, x in R`
C. `phi(x)=-x , x in (-oo,0]`
D. `phi (x)=x+|x|, x in R`

Answer

To solve the problem, we need to find the function \( \phi(x) \) that satisfies the equation:

\[ (\phi(x) - f(x))^2 + (\phi(x) - g(x))^2 = 0 \]

where \( f(x) = x \) and \( g(x) = |x| \).

### Step 1: Analyze the Equation

The equation states that the sum of two squares is equal to zero. The only way for a sum of squares to equal zero is if each square is individually zero. Therefore, we can set up the following equations:

\[ \phi(x) - f(x) = 0 \quad \text{and} \quad \phi(x) - g(x) = 0 \]

### Step 2: Solve for \( \phi(x) \)

From the first equation:

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

From the second equation:

\[ \phi(x) = g(x) = |x| \]

### Step 3: Determine the Conditions for Equality

For both equations to hold true simultaneously, we need:

\[ x = |x| \]

This condition holds true when \( x \geq 0 \) (i.e., \( x \) is non-negative).

### Step 4: Conclusion

Thus, we can conclude that:

- For \( x \geq 0 \), \( \phi(x) = x \).
- For \( x < 0 \), \( \phi(x) \) cannot equal both \( x \) and \( |x| \) simultaneously since \( x \) would be negative while \( |x| \) would be positive.

Therefore, the function \( \phi(x) \) that satisfies the given condition is:

\[ \phi(x) = x \quad \text{for } x \geq 0 \]

### Final Answer

The function \( \phi(x) \) is:

\[ \phi(x) = x \quad \text{for } x \geq 0 \]