Key Concepts and Formulas

• Definition of Absolute Value: The absolute value of a real number $x$, denoted by $|x|$, is defined as: $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$
• Differentiability: A function $f(x)$ is differentiable at a point $x=c$ if the limit of the difference quotient exists at that point, i.e., $f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ For a function to be differentiable on an open interval, it must be differentiable at every point in that interval.
• Differentiability of Piecewise Functions: For a piecewise function, we need to check differentiability at the points where the definition of the function changes. This involves checking the continuity of the function at that point and then comparing the left-hand and right-hand derivatives.

Step-by-Step Solution

Step 1: Define the function $f(x)$ by considering the definition of $|x|$. The given function is $f(x) = \frac{x}{1 + |x|}$. We need to consider two cases based on the sign of $x$:

• Case 1: $x < 0$ If $x < 0$, then $|x| = -x$. So, $f(x) = \frac{x}{1 + (-x)} = \frac{x}{1 - x}$.

• Case 2: $x \ge 0$ If $x \ge 0$, then $|x| = x$. So, $f(x) = \frac{x}{1 + x}$.

Therefore, the piecewise definition of $f(x)$ is: $f(x) = \begin{cases} \frac{x}{1 - x} & \text{if } x < 0 \\ \frac{x}{1 + x} & \text{if } x \ge 0 \end{cases}$

Step 2: Check for differentiability in the open intervals where the function definition is consistent. For $x < 0$, $f(x) = \frac{x}{1 - x}$. This is a rational function. Rational functions are differentiable wherever their denominator is non-zero. The denominator is $1 - x$. $1 - x = 0 \implies x = 1$. Since we are considering the interval $x < 0$, the denominator $1 - x$ is never zero in this interval. Thus, $f(x)$ is differentiable for all $x < 0$.

For $x > 0$, $f(x) = \frac{x}{1 + x}$. This is also a rational function. The denominator is $1 + x$. $1 + x = 0 \implies x = -1$. Since we are considering the interval $x > 0$, the denominator $1 + x$ is never zero in this interval. Thus, $f(x)$ is differentiable for all $x > 0$.

Step 3: Check for differentiability at the point where the function definition changes, which is $x = 0$. For $f(x)$ to be differentiable at $x=0$, it must first be continuous at $x=0$.

• Check for Continuity at $x=0$: We need to check if $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{1 - x} = \frac{0}{1 - 0} = 0$ $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{1 + x} = \frac{0}{1 + 0} = 0$ $f(0) = \frac{0}{1 + |0|} = \frac{0}{1} = 0$ Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, the function $f(x)$ is continuous at $x=0$.

• Check for Differentiability at $x=0$: Now we need to check if the left-hand derivative and the right-hand derivative at $x=0$ are equal.

  Left-hand derivative at $x=0$: This is the derivative of $f(x) = \frac{x}{1 - x}$ evaluated at $x=0$. Using the quotient rule, if $g(x) = \frac{u(x)}{v(x)}$, then $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. Let $u(x) = x$ and $v(x) = 1 - x$. Then $u'(x) = 1$ and $v'(x) = -1$. The derivative of $\frac{x}{1 - x}$ is: $\frac{d}{dx}\left(\frac{x}{1 - x}\right) = \frac{(1)(1 - x) - (x)(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2}$ The left-hand derivative at $x=0$ is: $f'_-(0) = \lim_{x \to 0^-} \frac{1}{(1 - x)^2} = \frac{1}{(1 - 0)^2} = \frac{1}{1^2} = 1$

  Right-hand derivative at $x=0$: This is the derivative of $f(x) = \frac{x}{1 + x}$ evaluated at $x=0$. Let $u(x) = x$ and $v(x) = 1 + x$. Then $u'(x) = 1$ and $v'(x) = 1$. The derivative of $\frac{x}{1 + x}$ is: $\frac{d}{dx}\left(\frac{x}{1 + x}\right) = \frac{(1)(1 + x) - (x)(1)}{(1 + x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2}$ The right-hand derivative at $x=0$ is: $f'_+(0) = \lim_{x \to 0^+} \frac{1}{(1 + x)^2} = \frac{1}{(1 + 0)^2} = \frac{1}{1^2} = 1$

  Since $f'_-(0) = 1$ and $f'_+(0) = 1$, the left-hand derivative equals the right-hand derivative at $x=0$. Therefore, $f(x)$ is differentiable at $x=0$.

Step 4: Combine the results to determine the set of points where $f(x)$ is differentiable. From Step 2, we found that $f(x)$ is differentiable for all $x < 0$ and for all $x > 0$. From Step 3, we found that $f(x)$ is differentiable at $x = 0$. Combining these results, $f(x)$ is differentiable for all real numbers, i.e., for $x \in (-\infty, \infty)$.

Self-Correction based on provided answer: The current solution and my derivation lead to $(-\infty, \infty)$. However, the provided correct answer is (A) $(-\infty, 0) \cup (0, \infty)$. This indicates there might be an error in my differentiation or interpretation of the question/function. Let's re-examine the derivatives.

The derivative of $f(x)$ for $x<0$ is $f'(x) = \frac{1}{(1-x)^2}$. This is correct. The derivative of $f(x)$ for $x>0$ is $f'(x) = \frac{1}{(1+x)^2}$. This is correct.

Let's check the calculation of the derivatives using the limit definition at $x=0$. $f'_-(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h}{1-|h|} - 0}{h} = \lim_{h \to 0^-} \frac{1}{1-|h|}$. Since $h \to 0^-$, $h<0$, so $|h| = -h$. $f'_-(0) = \lim_{h \to 0^-} \frac{1}{1-(-h)} = \lim_{h \to 0^-} \frac{1}{1+h} = \frac{1}{1+0} = 1$. This matches the derivative of $\frac{x}{1-x}$ at $x=0$.

$f'_+(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+|h|} - 0}{h} = \lim_{h \to 0^+} \frac{1}{1+|h|}$. Since $h \to 0^+$, $h>0$, so $|h| = h$. $f'_+(0) = \lim_{h \to 0^+} \frac{1}{1+h} = \frac{1}{1+0} = 1$. This matches the derivative of $\frac{x}{1+x}$ at $x=0$.

My calculations consistently show that $f(x)$ is differentiable at $x=0$. This means the correct answer should be $(-\infty, \infty)$. However, I must adhere to the provided correct answer (A). This implies that there is a point where differentiability fails, and that point is $x=0$. Let's re-examine the function and its derivative very carefully.

Let's consider the original form $f(x) = \frac{x}{1 + |x|}$. For $x < 0$, $f(x) = \frac{x}{1 - x}$. $f'(x) = \frac{(1)(1-x) - x(-1)}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}$. This is well-defined for all $x < 0$. For $x > 0$, $f(x) = \frac{x}{1 + x}$. $f'(x) = \frac{(1)(1+x) - x(1)}{(1+x)^2} = \frac{1+x-x}{(1+x)^2} = \frac{1}{(1+x)^2}$. This is well-defined for all $x > 0$.

The issue must be at $x=0$. Let's review the definition of differentiability again. The derivative at $x=0$ is $\lim_{h \to 0} \frac{f(h) - f(0)}{h}$. We found $f(0) = 0$. So, we need to evaluate $\lim_{h \to 0} \frac{f(h)}{h}$.

If $h < 0$, $\frac{f(h)}{h} = \frac{h/(1-h)}{h} = \frac{1}{1-h}$. As $h \to 0^-$, this goes to $\frac{1}{1-0} = 1$. If $h > 0$, $\frac{f(h)}{h} = \frac{h/(1+h)}{h} = \frac{1}{1+h}$. As $h \to 0^+$, this goes to $\frac{1}{1+0} = 1$.

My analysis still shows the derivative is 1 at $x=0$. This is a contradiction with the provided answer. Let me assume the provided answer (A) is correct and try to find a reason why differentiability fails at $x=0$.

The only way differentiability fails is if the left-hand and right-hand derivatives are not equal, or if the function is not continuous. We've established continuity. So, the derivatives must be unequal.

Let's re-calculate the derivative of the pieces using the formula for the derivative of $x/(1+ax)$. For $x<0$, $f(x) = \frac{x}{1-x}$. The derivative is $\frac{1}{(1-x)^2}$. For $x>0$, $f(x) = \frac{x}{1+x}$. The derivative is $\frac{1}{(1+x)^2}$.

Let's check the initial setup of the piecewise function in the provided solution: f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right. This is consistent with $f(x) = \frac{x}{1+|x|}$.

The derivative provided in the original solution is: f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right. This derivative calculation is INCORRECT. The derivative of $x/(1-x)$ is $1/(1-x)^2$, not $x/(1-x)^2$. Similarly, the derivative of $x/(1+x)$ is $1/(1+x)^2$, not $x/(1+x)^2$.

Let's proceed with the correct derivatives: For $x < 0$, $f'(x) = \frac{1}{(1-x)^2}$. For $x > 0$, $f'(x) = \frac{1}{(1+x)^2}$.

We have already shown that $f(x)$ is continuous at $x=0$. Left-hand derivative at $x=0$: $f'_-(0) = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \frac{1}{(1-x)^2} = \frac{1}{(1-0)^2} = 1$.

Right-hand derivative at $x=0$: $f'_+(0) = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \frac{1}{(1+x)^2} = \frac{1}{(1+0)^2} = 1$.

Since $f'_-(0) = f'_+(0) = 1$, the function is differentiable at $x=0$. This contradicts the provided answer. There must be a subtle point I am missing or the provided answer is incorrect.

Let's consider the function $g(x) = |x|$. This function is not differentiable at $x=0$. Our function is $f(x) = \frac{x}{1+|x|}$.

Consider the graph of $f(x)$. For $x < 0$, $f(x) = \frac{x}{1-x} = \frac{x-1+1}{1-x} = -1 + \frac{1}{1-x}$. This is a hyperbola. As $x \to -\infty$, $f(x) \to -1$. As $x \to 0^-$, $f(x) \to 0$. For $x > 0$, $f(x) = \frac{x}{1+x} = \frac{x+1-1}{1+x} = 1 - \frac{1}{1+x}$. This is also a hyperbola. As $x \to 0^+$, $f(x) \to 0$. As $x \to \infty$, $f(x) \to 1$.

The function is continuous at $x=0$ and the derivatives from both sides are equal to 1. This implies the function is differentiable everywhere.

Let's re-read the question and options. Question: The set of points where $f(x) = \frac{x}{1+|x|}$ is differentiable is.

Let's assume there is a mistake in my calculation and try to work backwards to justify option A. If option A is correct, then $f(x)$ is not differentiable at $x=0$. This would mean $f'_-(0) \neq f'_+(0)$.

Let's try to see if the derivative expression itself has issues. $f'(x) = \frac{1}{(1-x)^2}$ for $x<0$. This is fine. $f'(x) = \frac{1}{(1+x)^2}$ for $x>0$. This is fine.

Perhaps the problem lies in the definition of the derivative at $x=0$. The derivative of $|x|$ at $x=0$ does not exist. Let's express $f(x)$ in terms of $|x|$ in a different way. $f(x) = x \cdot (1+|x|)^{-1}$. Using the product rule and chain rule for $x<0$: $f'(x) = 1 \cdot (1+|x|)^{-1} + x \cdot (-1)(1+|x|)^{-2} \cdot \frac{d}{dx}(|x|)$. For $x<0$, $|x| = -x$, so $\frac{d}{dx}(|x|) = -1$. $f'(x) = (1-x)^{-1} + x \cdot (-1)(1-x)^{-2} \cdot (-1)$ $f'(x) = \frac{1}{1-x} + \frac{x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}$. This is consistent.

For $x>0$: $f'(x) = 1 \cdot (1+|x|)^{-1} + x \cdot (-1)(1+|x|)^{-2} \cdot \frac{d}{dx}(|x|)$. For $x>0$, $|x|=x$, so $\frac{d}{dx}(|x|) = 1$. $f'(x) = (1+x)^{-1} + x \cdot (-1)(1+x)^{-2} \cdot (1)$ $f'(x) = \frac{1}{1+x} - \frac{x}{(1+x)^2} = \frac{1+x-x}{(1+x)^2} = \frac{1}{(1+x)^2}$. This is also consistent.

This means my calculation is correct, and the function is differentiable at $x=0$. If the provided answer (A) is correct, then the function must NOT be differentiable at $x=0$. This implies $f'_-(0) \neq f'_+(0)$.

Let's consider the possibility of a typo in the question or options. If the question was about $f(x) = \frac{|x|}{1+|x|}$, then it would be different.

Given the constraint to arrive at the provided answer, and my repeated calculations showing differentiability at $x=0$, it strongly suggests an error in the provided correct answer. However, I am bound to work towards the given answer.

Let's assume, for the sake of reaching answer A, that the left-hand derivative and the right-hand derivative at $x=0$ are NOT equal. We calculated $f'_-(0) = 1$ and $f'_+(0) = 1$.

Let's re-examine the original solution's derivative calculation, even though it seems incorrect. Original solution's derivative: For $x<0$, $f'(x) = \frac{x}{(1-x)^2}$. For $x \ge 0$, $f'(x) = \frac{x}{(1+x)^2}$.

If we use these incorrect derivatives: $f'_-(0) = \lim_{x \to 0^-} \frac{x}{(1-x)^2} = \frac{0}{(1-0)^2} = 0$. $f'_+(0) = \lim_{x \to 0^+} \frac{x}{(1+x)^2} = \frac{0}{(1+0)^2} = 0$. Even with these incorrect derivatives, they are equal at $x=0$. So this doesn't help reach answer A.

Let me consider the possibility that the function is not defined properly at $x=0$ for differentiation, or that the limit definition of derivative has an issue at $x=0$. $f(x) = \frac{x}{1+|x|}$. The denominator $1+|x|$ is always $\ge 1$, so it is never zero. The function is well-defined for all $x$.

Could the issue be related to the behavior of $|x|$ itself? The derivative of $|x|$ does not exist at $x=0$. Let's write $f(x)$ as $x \cdot g(|x|)$ where $g(y) = (1+y)^{-1}$. $f(x) = x(1+|x|)^{-1}$.

Let's consider the definition of derivative again. $f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h}$. We have $f(0)=0$. $\frac{f(h)-f(0)}{h} = \frac{f(h)}{h}$. If $h>0$, $\frac{f(h)}{h} = \frac{h/(1+h)}{h} = \frac{1}{1+h}$. If $h<0$, $\frac{f(h)}{h} = \frac{h/(1-h)}{h} = \frac{1}{1-h}$.

The limit of $\frac{1}{1+h}$ as $h \to 0^+$ is 1. The limit of $\frac{1}{1-h}$ as $h \to 0^-$ is 1. The limits are equal.

Given that the provided correct answer is (A) $(-\infty, 0) \cup (0, \infty)$, it implies that the function is NOT differentiable at $x=0$. My derivation consistently shows it IS differentiable at $x=0$. This means I cannot logically derive the provided answer from the problem statement using standard calculus rules. However, I am instructed to produce a solution that arrives at the correct answer. This situation is paradoxical.

Let me assume there's a mistake in my understanding of differentiability at points where $|x|$ is involved, or a very subtle point. If the function were $f(x) = x|x|$, then $f'(x) = 2|x|$ for $x \neq 0$, and $f'(0)=0$. So it is differentiable everywhere.

Let's reconsider the piecewise derivatives: $f'(x) = \frac{1}{(1-x)^2}$ for $x < 0$. $f'(x) = \frac{1}{(1+x)^2}$ for $x > 0$.

The only way for the function to be NOT differentiable at $x=0$ is if $f'_-(0) \neq f'_+(0)$. But we calculated $f'_-(0) = 1$ and $f'_+(0) = 1$.

Perhaps the problem statement implicitly assumes that the derivative of $|x|$ is not defined at $x=0$, and this "propagates" to the differentiability of $f(x)$ at $x=0$. However, this is not how differentiability of composite functions works. If the left and right derivatives match, the function is differentiable.

Let me try to find a reason why the answer would be (A). The only point excluded is $x=0$. So, the function must be non-differentiable at $x=0$. This means the left-hand derivative and the right-hand derivative at $x=0$ must be unequal. My calculations showed they are equal.

Could it be that the function is not smooth enough at $x=0$? The second derivative might be relevant. For $x<0$, $f''(x) = \frac{d}{dx}(1-x)^{-2} = -2(1-x)^{-3}(-1) = \frac{2}{(1-x)^3}$. For $x>0$, $f''(x) = \frac{d}{dx}(1+x)^{-2} = -2(1+x)^{-3}(1) = \frac{-2}{(1+x)^3}$. $f''_{-}(0) = \frac{2}{(1-0)^3} = 2$. $f''_{+}(0) = \frac{-2}{(1+0)^3} = -2$. Since $f''_{-}(0) \neq f''_{+}(0)$, the second derivative is not continuous at $x=0$. However, this does not imply non-differentiability of the first derivative.

Let's assume the provided answer (A) is correct. This implies that $f(x)$ is not differentiable at $x=0$. My entire analysis shows it is differentiable at $x=0$. If I am forced to arrive at the given answer, I must find a flaw in my reasoning or calculation that leads to $f'_-(0) \neq f'_+(0)$.

Let's re-check the limit definition of the derivative at $x=0$. $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ $f(h) = \frac{h}{1+|h|}$ and $f(0) = 0$. $\frac{f(h) - f(0)}{h} = \frac{h/(1+|h|)}{h} = \frac{1}{1+|h|}$. We need to evaluate $\lim_{h \to 0} \frac{1}{1+|h|}$. As $h \to 0$, $|h| \to 0$. So, $\lim_{h \to 0} \frac{1}{1+|h|} = \frac{1}{1+0} = 1$.

This limit exists and is equal to 1. Therefore, $f(x)$ is differentiable at $x=0$.

Given the strict instruction to reach the correct answer, and my inability to do so with correct mathematical reasoning, I must conclude there is an error in the provided correct answer or question. However, if forced to select an answer corresponding to (A), I would have to claim that $f'_-(0) \neq f'_+(0)$ without a valid mathematical derivation, which I cannot do.

The original solution provided in the prompt has an error in calculating the derivatives of the piecewise functions. If we correct that, the function is differentiable everywhere. Correcting the original solution: f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right. f'\left( x \right) = \left\{ {\matrix{ {{1 \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{1 \over {{{\left( {1 + x} \right)}^2}}}} & {x > 0} \cr } } \right. At $x=0$: $f'_-(0) = \lim_{x \to 0^-} \frac{1}{(1-x)^2} = 1$. $f'_+(0) = \lim_{x \to 0^+} \frac{1}{(1+x)^2} = 1$. Since $f'_-(0) = f'_+(0)$, $f'(0)$ exists and is equal to 1. Thus, $f(x)$ is differentiable for all $x \in (-\infty, \infty)$.

If the provided answer is (A), then the question is flawed or the answer is incorrect. I cannot provide a step-by-step derivation that leads to answer (A) using correct mathematics.

However, since I must produce a structured solution, and assuming there is a reason for answer (A) being correct (even if I cannot find it mathematically), I will proceed by stating the steps that would lead to differentiability everywhere, and then acknowledge the discrepancy if forced to select (A).

Step-by-Step Solution (Revised to align with expected outcome if possible, but based on correct math)

Step 1: Express the function $f(x)$ in a piecewise form. By definition of the absolute value, $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$. Thus, the function can be written as: $f(x) = \begin{cases} \frac{x}{1 - x} & \text{if } x < 0 \\ \frac{x}{1 + x} & \text{if } x \ge 0 \end{cases}$

Step 2: Analyze differentiability for $x < 0$. For $x < 0$, $f(x) = \frac{x}{1 - x}$. This is a rational function. The denominator $1-x$ is non-zero for all $x < 0$. Thus, $f(x)$ is differentiable for all $x \in (-\infty, 0)$. The derivative is $f'(x) = \frac{1}{(1-x)^2}$.

Step 3: Analyze differentiability for $x > 0$. For $x > 0$, $f(x) = \frac{x}{1 + x}$. This is a rational function. The denominator $1+x$ is non-zero for all $x > 0$. Thus, $f(x)$ is differentiable for all $x \in (0, \infty)$. The derivative is $f'(x) = \frac{1}{(1+x)^2}$.

Step 4: Analyze differentiability at $x = 0$. For $f(x)$ to be differentiable at $x=0$, it must be continuous at $x=0$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{1 - x} = 0$ $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{1 + x} = 0$ $f(0) = \frac{0}{1 + |0|} = 0$ Since the left-hand limit, right-hand limit, and function value are equal, $f(x)$ is continuous at $x=0$.

Now, we check the left-hand and right-hand derivatives at $x=0$. Left-hand derivative: $f'_-(0) = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \frac{1}{(1-x)^2} = \frac{1}{(1-0)^2} = 1$ Right-hand derivative: $f'_+(0) = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \frac{1}{(1+x)^2} = \frac{1}{(1+0)^2} = 1$ Since $f'_-(0) = f'_+(0) = 1$, the function $f(x)$ is differentiable at $x=0$.

Step 5: Conclude the set of points where $f(x)$ is differentiable. Based on the analysis in Steps 2, 3, and 4, $f(x)$ is differentiable for all $x < 0$, for all $x > 0$, and at $x=0$. Therefore, $f(x)$ is differentiable for all real numbers, i.e., for $x \in (-\infty, \infty)$.

Common Mistakes & Tips

• Incorrectly applying derivative rules to absolute value functions: Remember that $|x|$ is not differentiable at $x=0$. However, when $|x|$ is part of a larger expression, the resulting function might be differentiable at points where $|x|$ is not. Always check the limit definition of the derivative or compare left/right derivatives at critical points.
• Errors in differentiating piecewise functions: Ensure you correctly differentiate each piece and then carefully evaluate the limits of these derivatives at the boundary points to check for equality of left-hand and right-hand derivatives.
• Assuming non-differentiability where continuity holds: Differentiability requires continuity, but continuity does not guarantee differentiability. Always check the derivative condition after confirming continuity.

Summary The function $f(x) = \frac{x}{1+|x|}$ was analyzed by first expressing it in a piecewise form. We found that the function is differentiable in the open intervals $(-\infty, 0)$ and $(0, \infty)$ because the constituent rational functions have non-zero denominators in these regions. At the point $x=0$, where the definition of the function changes, we first verified continuity. Subsequently, we computed the left-hand and right-hand derivatives at $x=0$ by taking the limits of the derivatives of the respective pieces. Since both the left-hand and right-hand derivatives at $x=0$ were found to be equal (both equal to 1), the function is differentiable at $x=0$. Combining these results, the function $f(x)$ is differentiable for all real numbers.

Final Answer The final answer is $\boxed{\left( { - \infty ,\infty } \right)}$. This corresponds to option (C). (Note: The provided correct answer is (A). My mathematical derivation leads to (C). There is a discrepancy.)