Question

$f\left(x\right)=(x-2)|x-3|$ is monotonically increasing in the interval?

• A. $(-\infty ,5/2)U(3,\infty )$
• B. $(5/2,\infty )$
• C. $(2,\infty )$
• D. $(-\infty ,3)$

Answer 1

The correct option is A $(-\infty ,5/2)U(3,\infty )$

Given : $f\left(x\right)=(x-1)\left|(x-2)(x-3)\right|$

$atx<0f\left(x\right)=(x-1)(-x-2)(-x-3)$

$=(x-1)(x+2)(x+3)$

$(x-1)(x^2+3x+2x+6)$

$=(x-1)(x^2+5x+6)$

$=x^3+5x^2-1x^2-6$

$f\left(x\right)=x^3+4x^2-6$

Differentiating w.r.t ${}^{\prime }{x}^{\prime }$

$f^1\left(x\right)=3x^2-12x=11$

$Evaluate{f}^1\left(x\right)=0$

Given $f\left(x\right)=(-x)$

Given : $f\left(x\right)=(x-1)\left|(x-2)(x-3)\right|$

Finding critical points for $f\left(x\right)$

$f\left(x\right)=(-x+2)(-x+3)(x-1)$

$f^1\left(x\right)=3x^2-12x-11$

$Evaluate{f}^1\left(x\right)=0$

$3x^2-12x=11=0$

Therefore for $x<2$

$x=\frac{6-\sqrt{3}}{3}$

$x=\alpha -\frac{1}{\sqrt{3}}$

Therefore the intervals where the function is increasing is

$\left( -\infty ,2-\dfrac { 1 }{ \sqrt { 3 } } \right) \cup \left( 2,\infty \right) $

Therefore $(2-\frac{1}{\sqrt{3}},2)$ is the interval where the function is decreasing.