Question

If a function $f\left(x\right)=\underset{0}{\overset{x}{\int }}\text{sgn}\left(t\right)(t^2-7t+6)dt$ is defined in $x\in (0,\infty ),$ then which of the following is/are correct?

• A. $f\left(x\right)$ has a local maxima at $x=1$
• B. $f\left(x\right)$ is increasing on $(0,1)$
• C. there exist $c\in (0,2)$ such that $f^{\prime }\left(c\right)=0$
• D. $f$ is decreasing on $(0,1)$

Answer 1

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

there exist $c\in (0,2)$ such that $f^{\prime }\left(c\right)=0$

Given : $f\left(x\right)=\underset{0}{\overset{x}{\int }}\text{sgn}\left(t\right)(t^2-7t+6)dt$
$\Rightarrow f^{\prime }\left(x\right)=\text{sgn}\left(x\right)(x^2-7x+6)$
For critical points : $f^{\prime }\left(x\right)=0$
$\Rightarrow \text{sgn}\left(x\right)(x^2-7x+6)=0\Rightarrow (x-1)(x-6)=0(\because \text{sgn}(x)>0\forall x>0)\Rightarrow x=1,6$
Sign scheme of $f^{\prime }\left(x\right):$