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

1 Answer
Apr 2, 2018

Answer:

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)