Question

If $f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^{t^2}(t-2)(t-3)dt$ 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 $f^{\prime \prime }\left(c\right)=0$
• D. $f$ has a local minimum at $x=3$

Answer 1

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

$f$ has a local minimum at $x=3$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}{e}^{t^2}(t-2)(t-3)dt$
$f^{\prime }\left(t\right)=e^{x^2}(x-2)(x-3)$

$f^{\prime }\left(x\right)<0\forall \in (2,3)$
So, $f\left(x\right)$ is decreasing on $(2,3)$
also at $x=2,f^{\prime }\left(x\right)$ changes its sign from positive to negative Hence, $x=2$ is point of maxima
At $x=3,f^{\prime }\left(x\right)$ changes its sign from negative to positive.
Hence, $x=3$ is point of minima.
Also $f^{\prime }\left(2\right)=f^{\prime }\left(3\right)=0$
So, from Rolle's Theorem there exist a point $c$ such that $f^{\prime \prime }\left(c\right)=0$