Question

Find the local minimum and maximum values of $f\left(x\right)=x^4-3x^3+3x^2-x$.

Answer 1

Given, $f\left(x\right)=x^4-3x^3+3x^2-x$
Thus, $f^{\prime }\left(x\right)=4x^3-9x^2+6x-1$
At a turning point, $f^{\prime }\left(x\right)=0$ gives
$4x^3-9x^2+6x-1=0$
$(x-1{)}^2(4x-1)=0\Rightarrow x=1,1,\frac{1}{4}$
When $x=1,f\left(1\right)=0$ and when $x=\frac{1}{4}$
$f\left(\frac{1}{4}\right)=\frac{-27}{256}$
Hence, the coordinates of the stationary points are
$(1,0)$ and $(\frac{1}{4},\frac{-27}{256})$
$f"\left(x\right)=12x^2-18x+6$
$=6(2x^2-3x+1)$
$=6(x-1)(2x-1)$
$x=1,f"\left(1\right)=0$

Thus the second derivative test gives no information about the extremum nature of $f$ at $x=1$.
When $x=\frac{1}{4},f"\left(\frac{1}{4}\right)=\frac{9}{4}>0$.

Hence $(\frac{1}{4},\frac{-27}{256})$ is a minimum point.