If `f(x)=max{x^(2)-4,|x-2|,|x-4|}` then:

Options

A. f (x) is continuous for all ` x in R`
B. f (x) is differentiable except at `x=(-1 pm sqrt33)/(2)`
C. f (x) has a critical point at x =2
D. f (x) has no maximum

Answer

To solve the problem, we need to analyze the function \( f(x) = \max\{x^2 - 4, |x - 2|, |x - 4|\} \).

### Step 1: Analyze Each Component of the Function
We will first analyze each of the three components of the function separately.

1. **For \( x^2 - 4 \)**:
- This is a parabola that opens upwards. The vertex is at \( (0, -4) \).
- The roots are at \( x = -2 \) and \( x = 2 \).

2. **For \( |x - 2| \)**:
- This is a V-shaped graph with a vertex at \( (2, 0) \).
- It is defined as:
- \( x - 2 \) for \( x \geq 2 \)
- \( -x + 2 \) for \( x < 2 \)

3. **For \( |x - 4| \)**:
- This is also a V-shaped graph with a vertex at \( (4, 0) \).
- It is defined as:
- \( x - 4 \) for \( x \geq 4 \)
- \( -x + 4 \) for \( x < 4 \)

### Step 2: Determine the Intervals for Each Function
Next, we will determine the intervals where each function is dominant.

- **For \( x < -2 \)**:
- \( x^2 - 4 \) is positive and increasing.
- \( |x - 2| = -x + 2 \) is decreasing.
- \( |x - 4| = -x + 4 \) is also decreasing.

Here, \( f(x) = x^2 - 4 \).

- **For \( -2 \leq x < 2 \)**:
- \( x^2 - 4 \) is negative.
- \( |x - 2| = -x + 2 \) is decreasing.
- \( |x - 4| = -x + 4 \) is decreasing.

Here, \( f(x) = |x - 2| \) or \( |x - 4| \) depending on the value of \( x \).

- **For \( 2 \leq x < 4 \)**:
- \( x^2 - 4 \) is increasing and positive.
- \( |x - 2| = x - 2 \) is increasing.
- \( |x - 4| = -x + 4 \) is decreasing.

Here, we need to compare \( x^2 - 4 \) and \( |x - 2| \).

- **For \( x \geq 4 \)**:
- All functions are increasing.
- Here, \( f(x) = x^2 - 4 \).

### Step 3: Find Points of Intersection
To find where the functions intersect, we set them equal to each other.

1. **Set \( x^2 - 4 = |x - 2| \)**:
- For \( x \geq 2 \): \( x^2 - 4 = x - 2 \) leads to \( x^2 - x - 2 = 0 \).
- For \( x < 2 \): \( x^2 - 4 = -x + 2 \) leads to \( x^2 + x - 6 = 0 \).

2. **Set \( x^2 - 4 = |x - 4| \)**:
- For \( x \geq 4 \): \( x^2 - 4 = x - 4 \) leads to \( x^2 - x = 0 \).
- For \( x < 4 \): \( x^2 - 4 = -x + 4 \) leads to \( x^2 + x - 8 = 0 \).

### Step 4: Analyze Continuity and Differentiability
- The function \( f(x) \) is continuous everywhere since it is defined as a maximum of continuous functions.
- To check differentiability, we need to analyze the points where the functions intersect, as these points may create sharp corners.

### Conclusion
- **Option A**: \( f(x) \) is continuous for all \( x \in \mathbb{R} \) - **True**.
- **Option B**: \( f(x) \) is differentiable except at the points of intersection found above - **True**.
- **Option C**: \( f(x) \) has no maxima - **True** since it is a parabola opening upwards.