Let $f(x)=x$ and $g(x)=|x|$ for all $x\in R,$ then the function $\phi(x)$ satisfying ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ is:

Q: Let $f(x)=x$ and $g(x)=|x|$ for all $x\in R,$ then the function $\phi(x)$ satisfying ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ is:

To determine the function $\phi(x)$ that satisfies the equation ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$, we need to analyze the given functions $f(x) = x$ and $g(x) = |x|$. The equation ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ implies that both $\phi(x) - f(x)$ and $\phi(x) - g(x)$ must be zero simultaneously. This is because the sum of two squares is zero only if each square is zero. Therefore, we have: 1. $\phi(x) - f(x) = 0$ which implies $\phi(x) = f(x) = x$. 2. $\phi(x) - g(x) = 0$ which implies $\phi(x) = g(x) = |x|$. For these two conditions to hold simultaneously, $\phi(x)$ must equal both $x$ and $|x|$. This is only possible when $x$ is non-negative, i.e., $x \geq 0$, because: - For $x \geq 0$, $|x| = x$, so both conditions $\phi(x) = x$ and $\phi(x) = |x|$ are satisfied. Therefore, $\phi(x) = x$ for $x \in [0, \infty)$. Now, let's examine why the other options are incorrect: 1. **Option 1: $\phi(x) = x + |x|$, $x \in R$** - For $x < 0$, $|x| = -x$, so $x + |x| = 0$, which does not satisfy $\phi(x) = x$ or $\phi(x) = |x|$. - For $x \geq 0$, $x + |x| = 2x$, which also does not satisfy $\phi(x) = x$ or $\phi(x) = |x|$. 2. **Option 2: $\phi(x) = x$, $x \in R$** - For $x < 0$, $|x| \neq x$, so $\phi(x) = x$ does not satisfy $\phi(x) = |x|$. 3. **Option 3: $\phi(x) = -x$, $x \in (-\infty, 0]$** - For $x \leq 0$, $-x \neq x$ and $-x \neq |x|$, so this does not satisfy either condition. Thus, the only option that satisfies the condition ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ for the specified domain is Option 4: $\phi(x) = x$, $x \in [0, \infty)$.

Question

Accepted Answer

To determine the function $\phi(x)$ that satisfies the equation ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$, we need to analyze the given functions $f(x) = x$ and $g(x) = |x|$.

The equation ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ implies that both $\phi(x) - f(x)$ and $\phi(x) - g(x)$ must be zero simultaneously. This is because the sum of two squares is zero only if each square is zero. Therefore, we have:

1. $\phi(x) - f(x) = 0$ which implies $\phi(x) = f(x) = x$.
2. $\phi(x) - g(x) = 0$ which implies $\phi(x) = g(x) = |x|$.

For these two conditions to hold simultaneously, $\phi(x)$ must equal both $x$ and $|x|$. This is only possible when $x$ is non-negative, i.e., $x \geq 0$, because:

- For $x \geq 0$, $|x| = x$, so both conditions $\phi(x) = x$ and $\phi(x) = |x|$ are satisfied.

Therefore, $\phi(x) = x$ for $x \in [0, \infty)$.

Now, let's examine why the other options are incorrect:

1. **Option 1: $\phi(x) = x + |x|$, $x \in R$**

- For $x < 0$, $|x| = -x$, so $x + |x| = 0$, which does not satisfy $\phi(x) = x$ or $\phi(x) = |x|$.
   - For $x \geq 0$, $x + |x| = 2x$, which also does not satisfy $\phi(x) = x$ or $\phi(x) = |x|$.

2. **Option 2: $\phi(x) = x$, $x \in R$**

- For $x < 0$, $|x| \neq x$, so $\phi(x) = x$ does not satisfy $\phi(x) = |x|$.

3. **Option 3: $\phi(x) = -x$, $x \in (-\infty, 0]$**

- For $x \leq 0$, $-x \neq x$ and $-x \neq |x|$, so this does not satisfy either condition.

Thus, the only option that satisfies the condition ${[\phi(x)-f(x)]}^{2}+{[\phi(x)-g(x)]}^{2}=0$ for the specified domain is Option 4: $\phi(x) = x$, $x \in [0, \infty)$.