Let `f(x) = {{:(int_(0)^(x) {5+|1-t|dt}, if x gt 2),(5x+1, if x le 2):}`, then

Options

A. f(x) is not continuous at x = 2
B. f(x) is continuous but not differentiable at x = 2
C. f(x) is differentiable everywhere
D. the right derivative of `f(x)` at x = 2 does not exist.

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

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

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

We need to evaluate \( \int_{0}^{x} (5 + |1 - t|) dt \). We will break this integral into two parts based on the value of \( t \):

1. For \( t \) in the interval \( [0, 1] \), \( |1 - t| = 1 - t \).
2. For \( t \) in the interval \( [1, x] \) (since \( x > 2 \)), \( |1 - t| = t - 1 \).

Thus, we can write:

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

### Step 2: Solve the first integral

Calculating the first integral:

\[ \int_{0}^{1} (5 + (1 - t)) dt = \int_{0}^{1} (6 - t) dt \]

Evaluating this integral:

\[ = \left[ 6t - \frac{t^2}{2} \right]_{0}^{1} = \left( 6(1) - \frac{1^2}{2} \right) - \left( 6(0) - \frac{0^2}{2} \right) = 6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2} \]

### Step 3: Solve the second integral

Now, calculating the second integral:

\[ \int_{1}^{x} (5 + (t - 1)) dt = \int_{1}^{x} (4 + t) dt \]

Evaluating this integral:

\[ = \left[ 4t + \frac{t^2}{2} \right]_{1}^{x} = \left( 4x + \frac{x^2}{2} \right) - \left( 4(1) + \frac{1^2}{2} \right) = 4x + \frac{x^2}{2} - (4 + \frac{1}{2}) = 4x + \frac{x^2}{2} - \frac{9}{2} \]

### Step 4: Combine both parts

Combining both integrals, we have:

\[ f(x) = \frac{11}{2} + \left( 4x + \frac{x^2}{2} - \frac{9}{2} \right) = 4x + \frac{x^2}{2} + 1 \]

Thus, for \( x > 2 \):

\[ f(x) = 4x + \frac{x^2}{2} + 1 \]

### Step 5: Check continuity at \( x = 2 \)

Now we need to check the continuity of \( f(x) \) at \( x = 2 \):

1. Calculate \( f(2) \):

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

2. Calculate \( \lim_{x \to 2^-} f(x) \):

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

3. Calculate \( \lim_{x \to 2^+} f(x) \):

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

Since all three values are equal, \( f(x) \) is continuous at \( x = 2 \).

### Step 6: Check differentiability at \( x = 2 \)

1. Calculate the left-hand derivative:

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

2. Calculate the right-hand derivative:

Using the derivative of \( f(x) = 4x + \frac{x^2}{2} + 1 \):

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

At \( x = 2 \):

\[ f'(2^+) = 4 + 2 = 6 \]

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

### Conclusion

Thus, we conclude that \( f(x) \) is continuous at \( x = 2 \) but not differentiable at \( x = 2 \).

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