Let a function $$f : [0, 5] \to R$$ be continuous, $$f(1) = 3$$ and $$F$$ be defined as:
$$F(x) = \int_1^x t^2 g(t) \; dt$$, where $$g(t) = \int_1^t f(u) \; du$$.
Then for the function $$F(x)$$, the point $$x = 1$$ is:

We are given a continuous function $$f:[0,5]\to\mathbb R$$ with the specific value $$f(1)=3.$$

Define another function

$$g(t)=\int_{1}^{t} f(u)\,du.$$

The Fundamental Theorem of Calculus states that if a function is continuous, then the derivative of its integral from a constant to the variable is the integrand itself. Hence we have

$$g'(t)=f(t).$$

Because the upper and lower limits of the integral coincide at $$t=1$$,

$$g(1)=\int_{1}^{1} f(u)\,du=0.$$

Next, the required function is

$$F(x)=\int_{1}^{x} t^{2}\,g(t)\,dt.$$

Again applying the Fundamental Theorem of Calculus, the first derivative of $$F$$ is

$$F'(x)=x^{2}g(x).$$

Now we evaluate this derivative at $$x=1$$:

$$F'(1)=1^{2}\,g(1)=1\cdot 0=0.$$

Because the first derivative vanishes, $$x=1$$ is a critical point.

To determine the nature of this critical point, we differentiate once more. Using the product rule, which states $$\frac{d}{dx}[u(x)v(x)]=u'(x)v(x)+u(x)v'(x),$$ we obtain

$$F''(x)=\frac{d}{dx}\bigl[x^{2}g(x)\bigr]=2x\,g(x)+x^{2}\,g'(x).$$

We already know $$g'(x)=f(x),$$ so

$$F''(x)=2x\,g(x)+x^{2}\,f(x).$$

Substituting $$x=1,$$ together with $$g(1)=0$$ and $$f(1)=3,$$ gives

$$F''(1)=2\cdot1\cdot0 + 1^{2}\cdot3 = 0 + 3 = 3.$$

The computed value $$F''(1)=3$$ is positive, which means the graph of $$F$$ is concave upward at $$x=1.$$ According to the second-derivative test, a positive second derivative at a critical point signifies a local minimum.

Hence, the correct answer is Option A.