Inflection point of (1, 6) for y = x^3 + ax^2 + bx + 1

[MarkSA]

Hello,

I have a problem.

For what values of the constants a and b is (1,6) a point of inflection on the curve y = x^3 + ax^2 + bx + 1?

I'm having some difficulty starting on this problem. What does a point of inflection tell me? I know it is where concavity changes from up to down or vice versa. Is it correct to say that would make f''(1) = 0 then? Since f''>0 is concave up and f''<0 is concave down.

Also, plugging in (1,6) to the equation gets:
6 = 1 + a + b + 1
4 = a + b

I'm not sure where to take it from here.

 

[Loren]

Yes. At the point of inflection, the second derivative of the original curve =0.

 

[MarkSA]

Okay, I think I made some progress on this. Went down to the 2nd derivative of the function and ended up with
y'' = 6x + 2a
So, 0 = 6 + 2a or a = -3.

Aha!

4 = a + b
4 = -3 + b
b = 7

So, the function that has a point of inflection at (1,6) is
y = x^3 - 3x^2 + 7x + 1

Does this look correct to you guys?