Question

If $f(x)=\int\limits_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$ for all $x \in(0, \infty),$ then

• A. $f$ has a local maximum at $x=2$
• B. $f$ is decreasing on (2,3)
• C. there exists some $c \in(0, \infty)$ such that $f^{\prime}(c)=0$
• D. $f$ has a local minimum at $x=3$

Answer 1

Answer: (A), (B), (C), (D)

$f'(x)=e^{x^{2}}(x-2)(x-3)$
Clearly, maxima at $x=2,$ minima at $x=3$ and
decreasing in $x \in(2,3)$.
$f'(x)=0$ for $x=2$ and $x=3 $ (Rolle's theorem)
so there exist $c \in(2,3)$ for which
$f''(c)=0$