Let `f: R to R` be a function defined by`f(x)="min" {x+1,|x|+1}.` Then, which of the following is true?

Options

A. `f(x)geq` for all `x epsilonR`
B. f(x) is not differentiable at x=1
C. f (x) is differentiable everywhere
D. f (x) is not differentiable at x = 0

Answer

To solve the problem, we need to analyze the function defined by \( f(x) = \min \{ x + 1, |x| + 1 \} \).

### Step 1: Understand the components of the function
We have two functions to consider:
1. \( g_1(x) = x + 1 \)
2. \( g_2(x) = |x| + 1 \)

### Step 2: Analyze \( g_1(x) = x + 1 \)
This is a linear function with a slope of 1. It intersects the y-axis at (0, 1) and the x-axis at (-1, 0). The graph is a straight line that increases indefinitely.

### Step 3: Analyze \( g_2(x) = |x| + 1 \)
The function \( |x| \) is V-shaped and is defined as:
- For \( x \geq 0 \), \( |x| = x \) so \( g_2(x) = x + 1 \)
- For \( x < 0 \), \( |x| = -x \) so \( g_2(x) = -x + 1 \)

The graph of \( g_2(x) \) intersects the y-axis at (0, 1) and has a vertex at (0, 1). For \( x < 0 \), it decreases with a slope of -1.

### Step 4: Find the intersection points
To find where \( g_1(x) \) and \( g_2(x) \) intersect, we set:
\[ x + 1 = |x| + 1 \]
This simplifies to:
\[ x = |x| \]
- For \( x \geq 0 \): \( x = x \) (always true)
- For \( x < 0 \): \( x = -x \) which gives \( x = 0 \)

Thus, the functions intersect at \( x = 0 \).

### Step 5: Determine the behavior of \( f(x) \)
- For \( x < 0 \): \( g_1(x) < g_2(x) \) because \( x + 1 < -x + 1 \). Therefore, \( f(x) = g_1(x) = x + 1 \).
- For \( x \geq 0 \): \( g_1(x) = g_2(x) \). Therefore, \( f(x) = g_1(x) = x + 1 \).

### Step 6: Combine the results
Thus, we can summarize:
\[ f(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases} \]
This means \( f(x) = x + 1 \) for all \( x \).

### Step 7: Check differentiability
Since \( f(x) = x + 1 \) is a linear function, it is differentiable everywhere.

### Conclusion
The function \( f(x) \) is differentiable everywhere.

### Final Answer
The correct option is that \( f(x) \) is differentiable everywhere.

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