Finding unknown constants

  • Thread starter MrJamesta
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  • #1
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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.
 

Answers and Replies

  • #2
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Tricky problem but if a function f(x) has a derivative at some specific point then what can you say about f(x)?
 
  • #3
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That f(x) is continuous at that specific point?
 
  • #4
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Yes, that's true, but I am thinking even more obviously than that?
 
  • #5
SammyS
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  • #8
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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.
 
  • #10
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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?
 
  • #11
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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.
 
  • #12
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To get back to the original question:
Oh, it's an x intercept.
Right. Can you use it to find another equation?
 
  • #13
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I found
0=a+b+c+d
4=8a+4b+2c+d
 
  • #14
SteamKing
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I found
0=a+b+c+d
4=8a+4b+2c+d
Now, use the other two equations and solve for the unknown coefficients.
 

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