If f (x) = \int_{0}^{x} e^{t^{2}} \left(\right. t - 2 \left.\right) \left(\right. t - 3 \left.\right) \textrm{ } dt \textrm{ }\textrm{ } for all x ∈ (0, ∞), then

Engineering

Mathematics

Maxima and Minima

Question

If f (x) = $\int_{0}^{x} e^{t^{2}} (t - 2) (t - 3) dt$ for all x ∈ (0, ∞), then

f has a local minimum at x = 3.

f is decreasing on (2, 3)

f has a local maximum at x = 2

there exists some c ∈ (0, ∞) such that f '' (c) = 0

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$f (x) = \int_{0}^{x} e^{t^{2}} (t - 2) (t - 3) dt$

f '(x) = $e^{x^{2}}$ (x – 2) (x – 3) = 0 $\Rightarrow$ x = 2, 3 are the critical point

f '(x) = 0 at 2 and 3 $\Rightarrow$ f "(x) will have atleast one root in (2, 3) by Rolle's Theorem.

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