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If $$f(x) = \int_{0}^{x} e^{t^2}(t - 2)(t - 3) dt$$ for all $$x \in (0, \infty)$$, then
f has a local maximum at x = 2
f is decreasing on (2, 3)
there exists some $$c \in (0, \infty)$$ such that $$f''(c) = 0$$
f has a local minimum at x = 3
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