Question

If $f\left(X\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt$ for all $x\in (0,\infty )$ , then

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

Answer 1

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

There exists some $c\in (0,\infty )$ such that $f^{\prime \prime }\left(c\right)=0$

$f\left(X\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt$

$f^{\prime }\left(x\right)=e^x(x-2)(x-3)$


$f\left(x\right)$ has a local maxima at $x=2$ and minima at $x=3$

$\Rightarrow f\left(x\right)$ is decreasing on $(2,3)$

$f^{\prime \prime }\left(x\right)=e^{x^2}(2x-5)+2x{e}^{x^2}(x^2-5x+6)$

$=e^{x^2}(2x^3-10x^2+8x-5)$

$\because f^{\prime }\left(0\right)=-5\text{and}{f}^{\prime \prime }\left(x\right)\to \infty asx\to \infty$

$\Rightarrow f^{\prime \prime }\left(x\right)$ is zero for some $x\in (0,\infty )$