Finding Unknown Constants for a Cubic Function with Given Derivatives

[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)

Homework Equations

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.

 

[paisiello2]

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

 

[MrJamesta]

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

 

[paisiello2]

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

 

[SammyS]

@MrJamesta

What is f(1) ?

 

[paisiello2]

You tell me.

 

[MrJamesta]

Oh, it's an x intercept.

 

[paisiello2]

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.

 

[Mark44]

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.

 

[paisiello2]

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?

 

[Mark44]

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.

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.

 

[mfb]

To get back to the original question:

Oh, it's an x intercept.

Right. Can you use it to find another equation?

 

[MrJamesta]

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

 

[SteamKing]

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

Now, use the other two equations and solve for the unknown coefficients.