Let a function f: `[0,5] rarr R` be continuous , f(1) =3 and F be definded as : `F(x)=int_(1)^(x)t^2 g(t)dt,` where `g(t) = int_(1)^(t)f(u)du.`
Then for the function F, the point x=1 is

Options

A. a point of local minima ,
B. not a critical point.
C. a point of indflection.
D. a point of local maxima

Answer

To solve the problem step by step, we will analyze the function \( F(x) \) and its derivatives to determine the nature of the point \( x = 1 \).

### Step 1: Define the functions
We are given:
- \( f: [0, 5] \to \mathbb{R} \) is continuous and \( f(1) = 3 \).
- \( g(t) = \int_1^t f(u) \, du \).
- \( F(x) = \int_1^x t^2 g(t) \, dt \).

### Step 2: Differentiate \( F(x) \)
To analyze the behavior of \( F(x) \) at \( x = 1 \), we need to find the first derivative \( F'(x) \) using the Fundamental Theorem of Calculus and the product rule:
\[ F'(x) = x^2 g(x). \]

### Step 3: Evaluate \( F'(1) \)
Now, we evaluate \( F'(1) \):
\[ F'(1) = 1^2 g(1) = g(1). \]
Next, we need to find \( g(1) \):
\[ g(1) = \int_1^1 f(u) \, du = 0. \]
Thus, we have:
\[ F'(1) = 0. \]

### Step 4: Differentiate \( F'(x) \) to find \( F''(x) \)
Since \( F'(1) = 0 \), we need to check the second derivative \( F''(x) \) to determine the nature of the critical point:
\[ F''(x) = 2x g(x) + x^2 g'(x). \]

### Step 5: Evaluate \( F''(1) \)
Now, we evaluate \( F''(1) \):
\[ F''(1) = 2 \cdot 1 \cdot g(1) + 1^2 g'(1) = 0 + g'(1) = g'(1). \]
To find \( g'(1) \), we differentiate \( g(t) \):
\[ g'(t) = f(t). \]
Thus,
\[ g'(1) = f(1) = 3. \]
Therefore,
\[ F''(1) = 3. \]

### Step 6: Determine the nature of the critical point
Since \( F'(1) = 0 \) and \( F''(1) > 0 \), this indicates that \( x = 1 \) is a local minimum for the function \( F(x) \).

### Conclusion
The point \( x = 1 \) is a local minimum for the function \( F \).