Properties and Extrema of Integral Defined Function

To analyze the properties of the given function, we apply the Fundamental Theorem of Calculus (Leibniz's Rule) to: $f(x) = \int_0^x e^{t^2} (t - 2)(t - 3) \, dt, \quad \text{for } x \in (0, \infty)$

Differentiating $f(x)$ with respect to $x$: $f'(x) = e^{x^2} (x - 2)(x - 3)$

Since the exponential function $e^{x^2} > 0$ for all $x \in (0, \infty)$, the sign of $f'(x)$ is entirely determined by the quadratic term $(x - 2)(x - 3)$:

• For $x \in (0, 2)$: $(x - 2) < 0$ and $(x - 3) < 0 \implies f'(x) > 0$.
• For $x \in (2, 3)$: $(x - 2) > 0$ and $(x - 3) < 0 \implies f'(x) < 0$.
• For $x \in (3, \infty)$: $(x - 2) > 0$ and $(x - 3) > 0 \implies f'(x) > 0$.

Now, let us evaluate each option:

1. Option (A): At $x = 2$, $f'(2) = 0$. As $x$ increases through $x = 2$, the derivative $f'(x)$ changes sign from positive to negative. Hence, by the first derivative test, $f$ attains a local maximum at $x = 2$. Therefore, (A) is correct.

2. Option (B): For all $x \in (2, 3)$, we have $f'(x) < 0$. Therefore, $f$ is strictly decreasing on $(2, 3)$. Therefore, (B) is correct.

3. Option (D): At $x = 3$, $f'(3) = 0$. As $x$ increases through $x = 3$, the derivative $f'(x)$ changes sign from negative to positive. Hence, by the first derivative test, $f$ attains a local minimum at $x = 3$. Therefore, (D) is correct.

4. Option (C): The derivative $f'(x)$ is continuous on $[2, 3]$ and differentiable on $(2, 3)$, with: $f'(2) = 0 \quad \text{and} \quad f'(3) = 0$ By Rolle's Theorem, there must exist at least one $c \in (2, 3) \subset (0, \infty)$ such that: $(f')'(c) = f''(c) = 0$ Therefore, (C) is correct.

Thus, the correct options are A, B, C, and D.