Find the values of a, b, c, and d so that the function f(x) = ax^3 + bx^2 + cx + d will have relative extrema at (−1, 1) and (−2, 4). How does one start to go about this? Apparently the answers are a = 6, b = 27, c = 36, and d = 16.

Question

mhchen · Answer

So you know how relative extremas have a slope of 0.

And the derivative of a function is the slope?

You can find the derivative of the function, plug in the x,y values, and solve for a,b,c,d.

justjm · Answer

Find f'(x) using power rule

$f'(x)=3ax^2+2bx+c$

Apply your understanding of how to find extrema on functions. You would set the derivative function to 0, and then find the x values. Then you would plug in those x values to the parent function to find the respective y value of those critical points. Mathematically, looks like this:

$3ax^2+2bx+c=0$
$x=-1, -2$

Plug in those x values to get two systems of equations

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

You are also given the original function with the extrema points. So plug in the points as well to get two more systems of equations

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

$a(-2)^3+b(-2)^2+c(-2)+d=4$

So now you have 4 systems of equations. Solve them through I guess?

$3a-2b+c=0$
$12a-4b+c=0$
$-a+b-c+d=1$
$-8a+4b-2c+d=4$

justjm · Answer

I'm not sure. Give a try and tell me if it works. I'm going through this concept at the moment as well

justjm · Answer

I think it should work but it'll be time-consuming. I'm not sure if there is a more efficient way. @Hero any thoughts

justjm · Answer

Yep it works. Just did it and got the values you told me.