Question

Let $f\left(x\right)=x^3-6x^2+9x+8$, then $f\left(x\right)$ is strictly decreasing in

• A. $(\infty ,1)$
• B. $(1,3)$
• C. $(3,\infty )$
• D. $(-\infty ,1)\cup (3,\infty )$

Answer 1

The correct option is B $(1,3)$

$f^{\prime }\left(x\right)=3x^2-12x+9=3(x^2-4x+3)=3(x-3)(x-1).$

If $f\left(x\right)$ is decreasing, then

$f^{\prime }\left(x\right)<0\Rightarrow 3(x-3)(x-1)<0$

$\Rightarrow (x-3)(x-1)<0\Rightarrow$ either $x-3>0,x-1<0$

$\Rightarrow x-3<0,x-1>0\Rightarrow$ either $x>3,x<1$

$\Rightarrow x<3,x>1\Rightarrow 3<x<1$ or $1<x<3$

$\Rightarrow x\in (1,3)[\because 3<x<1$ is not possible $]$