1. Given functions

We have $g(t)=\int_1^t f(u)\,du$ with $f$ continuous on $[0,5]$ and $f(1)=3$.

Also, $F(x)=\int_1^x t^2 g(t)\,dt.$

We need to determine the nature of the point $x=1$ for $F$.

2. Differentiate $F$ using Fundamental Theorem of Calculus

Since the integrand $t^2g(t)$ is continuous, $F'(x)=x^2 g(x).$

Now at $x=1$, $g(1)=\int_1^1 f(u)\,du=0.$ Therefore, $F'(1)=1^2\cdot g(1)=0.$

So $x=1$ is a critical point. Hence option D is false.

3. Find higher derivatives

Differentiate again: $F''(x)=\frac{d}{dx}\big(x^2 g(x)\big)=2xg(x)+x^2 g'(x).$

But $g'(x)=f(x)$ by the Fundamental Theorem of Calculus.

So, $F''(x)=2xg(x)+x^2 f(x).$

At $x=1$, $F''(1)=2(1)g(1)+(1)^2f(1)=0+3=3>0.$

Therefore, by the second derivative test, $x=1$ is a point of local minimum of $F$.

So option C is correct.

4. Check other options

• A: point of inflection — false, since $F''(1)=3\ne 0$ and is positive, so no sign change indication at $x=1$.
• B: local maxima — false, because $F''(1)>0$ implies local minimum, not maximum.
• C: local minima — true.
• D: not a critical point — false, because $F'(1)=0$.

5. Final answer

Hence, for $F$, the point $x=1$ is a point of local minima.