Key Concepts and Formulas

• Definition of Minimum: For any two real numbers $a$ and $b$, $\min\{a, b\}$ is the smaller of the two values. If $a \le b$, then $\min\{a, b\} = a$. If $b \le a$, then $\min\{a, b\} = b$.
• 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 of a Function: A function $f(x)$ is differentiable at a point $c$ if the limit of the difference quotient exists at $c$. Geometrically, this means the function has a well-defined tangent line at that point, and the graph does not have any sharp corners or breaks. A function is differentiable everywhere if it is differentiable at every point in its domain.
• Differentiability of Basic Functions: Polynomials are differentiable everywhere. The function $f(x) = x+1$ is a polynomial.

Step-by-Step Solution

Step 1: Analyze the function definition. We are given the function $f(x) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$. To understand this function better, we need to compare the two expressions inside the minimum function: $x+1$ and $|x|+1$.

Step 2: Compare the two expressions based on the sign of $x$. We need to consider two cases for $|x|$:

• Case 1: $x \ge 0$ If $x \ge 0$, then by the definition of absolute value, $|x| = x$. So, the two expressions inside the minimum become: $x+1$ and $x+1$. In this case, $x+1 = |x|+1$. Therefore, $f(x) = \min\{x+1, x+1\} = x+1$ for $x \ge 0$.

• Case 2: $x < 0$ If $x < 0$, then by the definition of absolute value, $|x| = -x$. So, the two expressions inside the minimum become: $x+1$ and $-x+1$. Now we need to compare $x+1$ and $-x+1$. Let's find when $x+1 \le -x+1$. $x \le -x$ $2x \le 0$ $x \le 0$ Since we are in the case where $x < 0$, the inequality $x+1 \le -x+1$ holds true for all $x < 0$. Therefore, $f(x) = \min\{x+1, -x+1\} = x+1$ for $x < 0$.

Step 3: Combine the cases to simplify the function definition. From Step 2, we found that:

• For $x \ge 0$, $f(x) = x+1$.
• For $x < 0$, $f(x) = x+1$.

Since both cases lead to $f(x) = x+1$, we can conclude that $f(x) = x+1$ for all $x \in \mathbb{R}$.

Step 4: Determine the differentiability of the simplified function. The simplified function is $f(x) = x+1$. This is a linear function, which is a type of polynomial. Polynomial functions are known to be differentiable everywhere on their domain, which is $\mathbb{R}$. The derivative of $f(x) = x+1$ is $f'(x) = 1$. Since the derivative exists for all real numbers, $f(x)$ is differentiable everywhere.

Step 5: Evaluate the given options.

• (A) $f(x)$ is differentiable everywhere. Our analysis in Step 4 confirms this.
• (B) $f(x)$ is not differentiable at $x=0$. This is false, as $f(x)$ is differentiable everywhere.
• (C) $f(x) > 1$ for all $x \in \mathbb{R}$. Since $f(x) = x+1$, for $x = -1$, $f(-1) = -1+1 = 0$, which is not greater than 1. So, this is false.
• (D) $f(x)$ is not differentiable at $x=1$. This is false, as $f(x)$ is differentiable everywhere.

Common Mistakes & Tips

• Misinterpreting the minimum function: A common mistake is to not carefully analyze which expression is smaller in the minimum function. Always break down the problem based on the conditions that change the value of the absolute value function (i.e., when the argument is positive or negative).
• Ignoring the domain: When dealing with piecewise functions or functions involving absolute values, ensure that the differentiability is checked at the points where the definition of the function changes. In this case, the function simplified to a single expression, making it easier.
• Confusing continuity and differentiability: A function must be continuous at a point to be differentiable at that point. However, a continuous function is not necessarily differentiable (e.g., $|x|$ at $x=0$). In this problem, the function turned out to be a simple linear function, which is both continuous and differentiable everywhere.

Summary

The given function $f(x) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$ was analyzed by considering the cases for the absolute value function. It was found that for $x \ge 0$, $|x|=x$, making $f(x) = \min\{x+1, x+1\} = x+1$. For $x < 0$, $|x|=-x$, and comparing $x+1$ and $-x+1$ showed that $x+1 \le -x+1$ for $x<0$. Thus, in all cases, $f(x) = x+1$. Since $f(x) = x+1$ is a linear function, it is differentiable everywhere. Therefore, option (A) is true.

The final answer is $\boxed{A}$.