Question

Let $f\left(x\right)=x^4-4x^3+6x^2-4x+1$, then

• A. $f$ increases on $[1,\infty )$
• B. $f$ decreases on $[1,\infty )$
• C. $f$ has a minimum at $x=-1$
• D. $f$ has neither maximum or minimum

Answer 1

Answer: (A), (C)

The correct options are
A $f$ increases on $[1,\infty )$
C $f$ has a minimum at $x=-1$

Given : $f\left(x\right)=x^4-4x^3+6x^2-4x+1$

Differentiate w.r.t x

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

$=4(x^3-3x^2+3x-1)$

$=4{(x-1)}^3$

$i)if{x}^{{}^{\prime }}\ge 1,f^{{}^{\prime }}\left(x\right)\ge 0$ (Therefore option A is correct)

ii) equate $f^{{}^{\prime }}\left(x\right)=0$

$4{(x-1)}^3=0$

$x=1$

iii) $f^{{}^{\prime \prime }}\left(x\right)=12{(x-1)}^2$

$f^{{}^{\prime \prime }}\left(x\right)=0[ifx=1]$

$f^{{}^{\prime \prime }}\left(x\right)$

$f^{{}^{\prime \prime \prime }}\left(x\right)=24(x-1)\left[if\right(x=1\left)\right]$

$f^{{}^{\prime \prime \prime \prime }}\left(x\right)=+24>0$

Therefor this becomes point of minima at $x=0\Rightarrow$ option C is zero.