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Finding the constants of a function

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose f(x)=ax^3+bx^2+cx+d has stationary points at (1,3) and (3,-3). Find the constants, a, b, c, d and sketch the graph of f(x).


    2. Relevant equations



    3. The attempt at a solution
    No idea where to start with this one. I know I have to eliminate 3 of the variables first and then use the found variable to solve for the rest, and that this involves using the stationary points and possibly the derivative of f(x), but I am not sure how to go about using all of these parts. If any one could be of assistance here it would be greatly appreciated, thanks in advance.
     
  2. jcsd
  3. Mar 25, 2009 #2
    "Stationary" points? As in relative extrema?
     
  4. Mar 26, 2009 #3
    Yupp, this is a relative extrema question.
     
  5. Mar 26, 2009 #4

    Mark44

    Staff: Mentor

    Stationary points are points where the derivative is zero. Start by calculating the derivative, f'(x).

    You have two points that are on the graph of the derivative. You are given that f'(1) = 3 and that f'(3) = -3. That will give you two equations in your three unknowns.

    What sort of a curve is the derivative function? Is there anything about its shape that you can use?
     
  6. Mar 26, 2009 #5
    I know it is a quadratic function, but other than that I don't know how to apply that knowledge to eliminate another variable.
     
  7. Mar 26, 2009 #6

    Mark44

    Staff: Mentor

    Start with what I said in post #4.
     
  8. Mar 26, 2009 #7
    That is what I have done. I found the derivative and created 2 equations using the points given (3=3a+2b+c and -3=27a+6b+c). From there I subracted the 2 equations to eliminate c and ended up with 24a+4b=-6, or b=-6a-(2/3). From that point I am unsure how to continue, as I know I have to eliminate either a or b now, but do not know how to go about it.
     
  9. Mar 27, 2009 #8

    Mark44

    Staff: Mentor

    Good work, but you have a small error. If 24a + 4b = -6, then b = -6a - (3/2).

    The derivative is f'(x) = 3ax2 + 2bx + c, which I'm sure you already know.
    Another way to write this is f'(x) = 3a(x2 + 2b/(3a) * x ) + c
    = 3a(x2 + 2b/(3a) * x + b2/(9a2) ) + c - (3ab2)/(9a2), by completing the square
    = 3a(x + b/(3a))2 + (9a2c - 3ab2)/(9a2)

    This is a parabola, and the low point (assuming a > 0) happens when x = ?
    And at that x value, y = ? This also gets you the axis of symmetry, which you might be able to use to get some more points on the graph of the derivative.

    I haven't worked it any farther than this, but I hope this gives you some ideas.
     
  10. Mar 27, 2009 #9
    Thanks for that correction Mark44, I caught it when I re-did it after. So I can see I now have a new point of (-b/3a, (9a^2c-3ab^2)/9a^2). Would I just plug this into the original function to get a new function, then sub the b=-6a-(3/2) into this and solve? That's really all I can see doing with this question, because frankly this one is stumping me :confused:
     
  11. Mar 27, 2009 #10

    Mark44

    Staff: Mentor

    The new point, (-b/3a, (9a^2c-3ab^2)/9a^2), is a point on the derivative, f'(x), whose graph is a parabola. We don't know yet if it opens up or down, but we can probably determine that from the other points on the parabola's graph, (1, 3) and (3, -3). In any case, the axis of symmetry for the parabola is x = -b/(3a), and you already have an equation for b in terms of a, so you should be able to use this to evaluate -b/(3a), and get a third point on the parabola. From this, I think you can find c, and we'll go from there.

    Havfind the vertex of the parabola
     
  12. Mar 27, 2009 #11
    Hmm by plugging in b into the axis of symmetry I get x=(6a+(2/3))/3a. I don't know what to do next; this new point gives me a value for x defined in terms of a, but how can I apply it? I had the idea to plug it into the derivative along with the value I found for b and the equation I found for y to try to solve for a, but this did not get me anywhere (not to mention seemed like too much work for this question :tongue: )
     
  13. Mar 28, 2009 #12

    Mark44

    Staff: Mentor

    Emethyst,
    It appears that I have led you astray, for which I apologize. The problem is much simpler than I have made it.

    Here's where we need to go.

    f(x) = ax3 + bx2 + cx + d
    f'(x) = 3ax2 +2bx +c

    Stationary points: (1, 3) and (3, -3)
    My mistake was saying that these were points on the derivative. They're not; they are points on the graph of the function.

    At stationary point, the derivative is 0, which means that f'(1) = 0 and f'(3) = 0.

    So, you have:
    f(1) = a + b + c + d = 3
    f(3) = 27a + 9b + 3c + d = -3

    f'(1) = 3a + 2b + c = 0
    f'(3) = 27a + 6b + c = 0

    Now you have four equations in four unknowns, so you should be able to solve this system and get your solution.

    Mark
     
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