Let `f(x)` be a periodic function with period `int_0^x f(t+n) dt 3 and f(-2/3)=7 and g(x) =` .where `n=3k, k in N`. Then `g'(7/3) =`

Options

A. `-2/3`
B. 7
C. -7
D. `7/3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the function g(x)
The function \( g(x) \) is defined as:
\[ g(x) = \int_0^x f(t + n) \, dt \]
where \( n = 3k \) for \( k \in \mathbb{N} \).

### Step 2: Differentiate g(x)
To find \( g'(x) \), we apply the Fundamental Theorem of Calculus:
\[ g'(x) = \frac{d}{dx} \left( \int_0^x f(t + n) \, dt \right) = f(x + n) \]
This is because the derivative of the integral with respect to its upper limit gives us the integrand evaluated at that limit.

### Step 3: Substitute n
Since \( n = 3k \), we can write:
\[ g'(x) = f(x + 3k) \]

### Step 4: Use the periodicity of f(x)
Given that \( f(x) \) is periodic with a period of 3, we can simplify \( f(x + 3k) \):
\[ g'(x) = f(x + 3k) = f(x) \]
This is because adding any multiple of the period (3) to \( x \) does not change the value of the periodic function.

### Step 5: Evaluate g'(7/3)
Now we need to find \( g'(7/3) \):
\[ g'(7/3) = f(7/3) \]

### Step 6: Relate 7/3 to the known value of f
We know from the problem statement that \( f(-2/3) = 7 \). We can relate \( 7/3 \) to \( -2/3 \):
\[ 7/3 = 3 - 2/3 \]
Since \( f(x) \) is periodic with period 3, we have:
\[ f(7/3) = f(3 - 2/3) = f(-2/3) \]

### Step 7: Substitute the known value
Since \( f(-2/3) = 7 \), we can conclude:
\[ f(7/3) = 7 \]

### Final Result
Thus, we find:
\[ g'(7/3) = 7 \]

### Summary
The value of \( g'(7/3) \) is \( 7 \).

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