Question

a challenge if"Y"=F(x) represents third degree polynomial function such that: F"(x)<0 when X<(-2/3) and f"(x)>0 when X>(-2/3) and the curve of the function passes by (1,6) and there exists a critical point at (-1,2) find the equation of this curve

Answer 1

is the answer: x^3 + 2 x^2 + x + 2 ???

Answer 2

@hartnm the answer is right but you have to show me the idea please

Answer 3

go on hartnn :)

Answer 4

let f(x)=a x^3 + b x^2 + cx + d
since,F"(x)<0 when X<(-2/3) and f"(x)>0 when X>(-2/3),
F"(x) must be 0 for X=-2/3
F"(x)=6ax+2b=0 for x=-2/3
so we get b=2a----->'W'
since the critical point exist on the curve,you have two points on the curve f(x)
so u get two equations like this
-a+b-c+d=2------>'P'
a+b+c+d=6------->'Q'
now critical point is at (-1,2) implies 1st derivative of f(x) is zero at that point
so 3a x^2 +2bx+c=0 at x=-1 gives
3a-2b+c=0------>'Z'
you have 4 equations 4 unknowns, very easily u can find them
ask if u need help in that....
@hamidic