let f(x)= ∫^x₀ 5+|1-t| dt, if x>2 5x+1, if x

1. discontinous at x=2
2. not differentiable at x=2
3. continuous at x=2
4. diffrentiable at x=2

To analyze the function \( f(x) \) defined piecewise, we need to break it down into its components and examine its behavior at the point of interest, which is \( x = 2 \). The function is defined as follows:

• For \( x \leq 2 \): \( f(x) = \int_0^x (5 + |1 - t|) \, dt \)
• For \( x > 2 \): \( f(x) = 5x + 1 \)

Step 1: Evaluating Continuity at \( x = 2 \)

To determine if \( f(x) \) is continuous at \( x = 2 \), we need to check the following:

• The limit of \( f(x) \) as \( x \) approaches 2 from the left, \( f(2^-) \).
• The limit of \( f(x) \) as \( x \) approaches 2 from the right, \( f(2^+) \).
• The value of \( f(2) \).

Calculating \( f(2) \)

First, we find \( f(2) \) using the integral definition:

\( f(2) = \int_0^2 (5 + |1 - t|) \, dt \)

To evaluate this integral, we need to consider the expression \( |1 - t| \). For \( t \) in the interval [0, 2], we have:

• For \( t < 1 \), \( |1 - t| = 1 - t \)
• For \( t \geq 1 \), \( |1 - t| = t - 1 \)

Thus, we can split the integral:

\( f(2) = \int_0^1 (5 + (1 - t)) \, dt + \int_1^2 (5 + (t - 1)) \, dt \)

Calculating each part:

\( \int_0^1 (6 - t) \, dt = \left[ 6t - \frac{t^2}{2} \right]_0^1 = 6 - \frac{1}{2} = \frac{11}{2} \)

\( \int_1^2 (4 + t) \, dt = \left[ 4t + \frac{t^2}{2} \right]_1^2 = (8 + 2) - (4 + 0.5) = 10 - 4.5 = 5.5 \)

Combining these results gives:

\( f(2) = \frac{11}{2} + 5.5 = \frac{11}{2} + \frac{11}{2} = 11 \)

Finding \( f(2^-) \) and \( f(2^+) \)

Next, we calculate the left-hand limit:

\( f(2^-) = f(2) = 11 \)

Now, for the right-hand limit:

\( f(2^+) = 5(2) + 1 = 10 + 1 = 11 \)

Step 2: Conclusion on Continuity

Since \( f(2^-) = f(2) = f(2^+) = 11 \), we conclude that \( f(x) \) is continuous at \( x = 2 \).

Step 3: Checking Differentiability at \( x = 2 \)

To determine if \( f(x) \) is differentiable at \( x = 2 \), we need to check the derivatives from both sides:

Finding \( f'(2^-) \)

Using the Fundamental Theorem of Calculus, we differentiate the integral for \( x < 2 \):

\( f'(x) = 5 + |1 - x| \)

At \( x = 2 \), we have:

\( f'(2^-) = 5 + |1 - 2| = 5 + 1 = 6 \)

Finding \( f'(2^+) \)

For \( x > 2 \), the derivative is simply:

\( f'(x) = 5 \)

At \( x = 2 \), we have:

\( f'(2^+) = 5 \)

Final Thoughts on Differentiability

Since \( f'(2^-) = 6 \) and \( f'(2^+) = 5 \), the left-hand and right-hand derivatives are not equal. Therefore, \( f(x) \) is not differentiable at \( x = 2 \).

In summary, the function \( f(x) \) is continuous at \( x = 2 \) but not differentiable at that point. This analysis highlights the importance of checking both continuity and differentiability separately, especially at points where the definition of the function changes.