$$\begin{aligned} & \text { Given, } f(x)=\int_0^x e^{t^2}(t-2)(t-3) d t, x \in(0, \infty) \\ & \Rightarrow \quad f^{\prime}(x)=e^{x^2}(x-2)(x-3) \end{aligned}$$

Here, $$f^{\prime}(x)$$ changes its sign $$+v e$$ to $$-v e$$ about $$x=2$$ and $$f^{\prime}(x)$$ changes its sign $$-v \mathrm{e}$$ to $$+v \mathrm{e}$$ about $$x=3$$

Hence, $$x=2$$ is the point of local maxima and $$x=3$$ is the point of local minima

$$\because f^{\prime}(x)<0$$ for $$x \in(2,3)$$

$$\therefore f(x)$$ is decreasing on $$x \in(2,3)$$

$$\because \quad f^{\prime}(x)$$ is continuous and differentiable for all $$x(0, \infty)$$ and $$f^{\prime}(2)=f^{\prime}(3)=0$$

$$\therefore$$ According to Rolle's theorem, $$f^{\prime \prime}(c)=0$$ must have at least one root $$\in(2,3)$$