# What are the extrema of y = x^4 - 3x^3 + 3x^2 - x?

Apr 2, 2018

the minima is $\left(\frac{1}{4} , - \frac{27}{256}\right)$ and the maxima is (1,0)

#### Explanation:

$y = {x}^{4} - 3 {x}^{3} + 3 {x}^{2} - x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 9 {x}^{2} + 6 x - 1$
For stationary points, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$4 {x}^{3} - 9 {x}^{2} + 6 x - 1$=0
$\left(x - 1\right) \left(4 {x}^{2} - 5 x + 1\right) = 0$
${\left(x - 1\right)}^{2} \left(4 x - 1\right) = 0$
$x = 1 \mathmr{and} x = \frac{1}{4}$

${d}^{2} \frac{y}{\mathrm{dx}} ^ 2$= $12 {x}^{2} - 18 x + 6$
Testing x=1
${d}^{2} \frac{y}{\mathrm{dx}} ^ 2$ = 0
therefore, possible horizontal point of inflexion (in this question, you don't need to find whether it is a horizontal point of inflexion)

Testing x=$\frac{1}{4}$
${d}^{2} \frac{y}{\mathrm{dx}} ^ 2$= $\frac{9}{4}$ >0
Therefore, minimum and concave up at x=$\frac{1}{4}$

Now, finding the x-intercepts,
let y=0
$\left({x}^{3} - x\right) \left(x - 3\right) = 0$
$x \left({x}^{2} - 1\right) \left(x - 3\right) = 0$
$x = 0 , \pm 1 , 3$

finding y-intercepts, let x=0
y=0 (0,0)
graph{x^4-3x^3+3x^2-x [-10, 10, -5, 5]}

From the graph, you can see that the minima is $\left(\frac{1}{4} , - \frac{27}{256}\right)$ and the maxima is (1,0)