Asked by Drake
If the function y=(x^2−2x)^2+6x^2−12x
has the minimum value b at x=a, what is a−b?
has the minimum value b at x=a, what is a−b?
Answers
Answered by
Anonymous
y=x^4-4x^3+4x^2+6x^2-12x
=x^4-4x^3+10x^2-12x
dy/dx=4x^3-12x^2+20x-12
Equate to 0, x^3-3x^2+5x-3=0
Factorising it we get (x-1)(x^2-2x+3)
x=1 or 1+-(-2)^1/2. Second root is imaginary hence x=1.
D2y/dx2=12x^2-24x+20. At x=1 it is=8 i.e.+ve hence y is min @ x=1=a, and
y min=1+4-4+6-12=-5=b
Hence a-b=1-(-5)=6
=x^4-4x^3+10x^2-12x
dy/dx=4x^3-12x^2+20x-12
Equate to 0, x^3-3x^2+5x-3=0
Factorising it we get (x-1)(x^2-2x+3)
x=1 or 1+-(-2)^1/2. Second root is imaginary hence x=1.
D2y/dx2=12x^2-24x+20. At x=1 it is=8 i.e.+ve hence y is min @ x=1=a, and
y min=1+4-4+6-12=-5=b
Hence a-b=1-(-5)=6
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