Question

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

Answer 1

the minima is `(1/4,-27/256)` and the maxima is (1,0)

Explanation:

`y=x^4-3x^3+3x^2-x`
`dy/dx = 4x^3-9x^2+6x-1`
For stationary points, `dy/dx=0`
`4x^3-9x^2+6x-1`=0
`(x-1)(4x^2-5x+1)=0`
`(x-1)^2(4x-1)=0`
`x=1 or x=1/4`

`d^2y/dx^2`= `12x^2-18x+6`
Testing x=1
`d^2y/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=`1/4`
`d^2y/dx^2`= `9/4` >0
Therefore, minimum and concave up at x=`1/4`

Now, finding the x-intercepts,
let y=0
`(x^3-x)(x-3)=0`
`x(x^2-1)(x-3)=0`
`x=0,+-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 `(1/4,-27/256)` and the maxima is (1,0)