THE FUNCTION Y=X^4+BX^2+8X+1 HAS A HORIZINTAL TANGENT

Question

AND A POINT OF INFLECTION FOR THE SAME VALUE OF X WHAT MUST BE THE VALUE OF B

Answer 1

dy/dx=4x^3+2bx+8. The tangent is horizontal when dy/dx=0: 4x^3+2bx+8=0; 2x^3+bx+4=0.

Differentiating again we get 12x^2+2b=0 at a point of inflection or undulation: 6x^2+b=0, making 6x^2=-b, so b<0 and x=±√(-b/6). Therefore x^2=-b/6 and x^3=-b√(-b/6)/6 and -b√(-b/6)/3+b√(-b/6)+4=(-1/3+1)b√(-b/6)+4=0.

So, 2b√(-b/6)/3=-4, 4b^2(-b/6)/9=16, -4b^3/6=144, b^3=-6*36=-216 so b=-6, making x=1 (not x=-1).

(Also, the function has a minimum at x=-2 when y'=0 (horizontal tangent). y=x^4-6x^2+8x+1.)