# Finding unknown constants

• MrJamesta

## Homework Statement

determine the constants a,b,c, and d so that the function f(x)=ax^3+bx^2+cx+d has its first derivative equal to 4 at the point (1,0) and its second derivative equal to 5 at the point (2,4)

## The Attempt at a Solution

I found the first and second derivative
f'(x)=3ax^2+2bx+c
f''(x)=6ax+2b

I set
5=12a+2b
I also set
4=3a+2b+c
I get stuck trying to find the two different variables when working with the second derivative first. I know I have to substitute for the unknowns, but I don't know where to start.

Tricky problem but if a function f(x) has a derivative at some specific point then what can you say about f(x)?

That f(x) is continuous at that specific point?

Yes, that's true, but I am thinking even more obviously than that?

You tell me.

Oh, it's an x intercept.

No, it's not.

f(1)=?

You have 4 unknowns so you need 4 equations to solve this. You came up with two of them. What are the other two? It's pretty obvious.

What is f(1) ?
Oh, it's an x intercept.

No, it's not.
Just to be clear, and to reduce confusion on the part of the OP, 1 is an x-intercept.

Well, what is an x-intercept? I think it is the value of the function when x=0. But f(1) is the value of the function when x=1. So it is not an x-intercept.

Regardless, it is irrelevant to solve the problem. What are the other two equations?

Well, what is an x-intercept? I think it is the value of the function when x=0.
No, that's the y-intercept, a point on the y-axis.
paisiello2 said:
But f(1) is the value of the function when x=1. So it is not an x-intercept.

Regardless, it is irrelevant to solve the problem.
But your reply to the OP was incorrect, possibly steering him/her in the wrong direction.

To get back to the original question:
Oh, it's an x intercept.
Right. Can you use it to find another equation?

I found
0=a+b+c+d
4=8a+4b+2c+d

I found
0=a+b+c+d
4=8a+4b+2c+d
Now, use the other two equations and solve for the unknown coefficients.