Let `f(x) = {{:(int_(0)^(x){5+|1-t|}dt",","if",x gt 2),(5x + 1",","if",x le 2):}`
Test `f(x)` for continuity and differentiability for all real x.

Answer

To solve the problem, we need to analyze the piecewise function defined as:

\[ f(x) = \begin{cases} \int_{0}^{x} (5 + |1 - t|) dt & \text{if } x > 2 \\ 5x + 1 & \text{if } x \leq 2 \end{cases} \]

### Step 1: Simplify the integral for \( x > 2 \)

For \( x > 2 \), we need to simplify the integral:

\[ |1 - t| = \begin{cases} 1 - t & \text{if } t < 1 \\ t - 1 & \text{if } t \geq 1 \end{cases} \]

Since \( x > 2 \), we can split the integral at \( t = 1 \):

\[ \int_{0}^{x} (5 + |1 - t|) dt = \int_{0}^{1} (5 + (1 - t)) dt + \int_{1}^{x} (5 + (t - 1)) dt \]

Calculating the first integral:

\[ \int_{0}^{1} (5 + (1 - t)) dt = \int_{0}^{1} (6 - t) dt = [6t - \frac{t^2}{2}]_{0}^{1} = (6 \cdot 1 - \frac{1^2}{2}) - (0) = 6 - 0.5 = 5.5 \]

Calculating the second integral:

\[ \int_{1}^{x} (5 + (t - 1)) dt = \int_{1}^{x} (t + 4) dt = [\frac{t^2}{2} + 4t]_{1}^{x} = \left(\frac{x^2}{2} + 4x\right) - \left(\frac{1^2}{2} + 4 \cdot 1\right) = \left(\frac{x^2}{2} + 4x\right) - \left(0.5 + 4\right) = \frac{x^2}{2} + 4x - 4.5 \]

Combining both parts, we have:

\[ f(x) = 5.5 + \left(\frac{x^2}{2} + 4x - 4.5\right) = \frac{x^2}{2} + 4x + 1 \]

### Step 2: Rewrite \( f(x) \)

Now we can rewrite \( f(x) \) as:

\[ f(x) = \begin{cases} \frac{x^2}{2} + 4x + 1 & \text{if } x > 2 \\ 5x + 1 & \text{if } x \leq 2 \end{cases} \]

### Step 3: Check continuity at \( x = 2 \)

To check continuity at \( x = 2 \), we need to find:

1. \( f(2) \)
2. The left-hand limit \( \lim_{x \to 2^-} f(x) \)
3. The right-hand limit \( \lim_{x \to 2^+} f(x) \)

Calculating \( f(2) \):

\[ f(2) = 5 \cdot 2 + 1 = 10 + 1 = 11 \]

Calculating the left-hand limit:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (5x + 1) = 5 \cdot 2 + 1 = 11 \]

Calculating the right-hand limit:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2} \left(\frac{x^2}{2} + 4x + 1\right) = \frac{2^2}{2} + 4 \cdot 2 + 1 = 2 + 8 + 1 = 11 \]

Since both limits and \( f(2) \) are equal, \( f(x) \) is continuous at \( x = 2 \).

### Step 4: Check differentiability at \( x = 2 \)

To check differentiability at \( x = 2 \), we need to find the left-hand derivative and the right-hand derivative.

Calculating the left-hand derivative:

\[ f'(x) = 5 \quad \text{for } x < 2 \]

Calculating the right-hand derivative:

\[ f'(x) = x + 4 \quad \text{for } x > 2 \]
Thus,
\[ f'(2) = 2 + 4 = 6 \]

Since the left-hand derivative (5) does not equal the right-hand derivative (6), \( f(x) \) is not differentiable at \( x = 2 \).

### Conclusion

- \( f(x) \) is continuous for all \( x \) in \( \mathbb{R} \).
- \( f(x) \) is differentiable for all \( x \) in \( \mathbb{R} \) except at \( x = 2 \).