# Math Help - Relative Extrema

1. ## Relative Extrema

I have to find a, b, c, and d such that the function defined by :
f(x) = $ax^3+bx^2+cx+d$
will have a relative extrema at points (1,2) and (2,3).

From the given critical points, I am able to know that when x=1 or x=2, f'(x)= $3ax^2+2bx+c$ should be equal to zero.

Therefore f'(x) should have factors like (x-1) and (x-2) or in simplified form $x^2-3x+2$. Now my problem is that how should I relate the two equations of f'(x) for me to be able to solve for the said unknowns?

2. You the coefficients of $f'(x)$ should have the ratio 1:-3:2. So $3a:2b:c = 1:-3:2$, which gives $\frac{3a}{2b} = \frac{1}{-3}$ and $\frac{2b}{c} = \frac{-3}{2}$. You get two more equations by plugging the points (1,2) and (2,3) into the function. So you get a system of 4 linear equations with 4 unknowns.

3. Originally Posted by deltaX
I have to find a, b, c, and d such that the function defined by :
f(x) = $ax^3+bx^2+cx+d$
will have a relative extrema at points (1,2) and (2,3).

From the given critical points, I am able to know that when x=1 or x=2, f'(x)= $3ax^2+2bx+c$ should be equal to zero.

Therefore f'(x) should have factors like (x-1) and (x-2) or in simplified form $x^2-3x+2$. Now my problem is that how should I relate the two equations of f'(x) for me to be able to solve for the said unknowns?
note that $f(1) = 2$

$a(1^3) + b(1^2) + c(1) + d = 2$

and $f(2) = 3$

$a(2^3) + b(2^2) + c(2) + d = 3$

$f'(1) = 0$

$3a(1^2) + 2b(1) + c = 0$

$f'(2) = 0$

$3a(2^2) + 2b(2) + c = 0$

four equations ... four unknown coefficients.

solve the system.