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Homework Help: Relative Minimum Problem

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f be a function given by f(x)= x^3 - 5x^2 + 3x + K, where k is a constant. Find the value of k for which f has 11 as it's relative minimum.

    2. Relevant equations

    f(x)= x^3 - 5x^2 + 3x + K

    3. The attempt at a solution
    First, I found the derivative of f(x), found the critcal numbers, and then tested for relative minimum. There was a relative minimum at x=3. Now I'm not quite sure what to do so that 11 is a relative min.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 11, 2008 #2

    HallsofIvy

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    Well, if the minimum is AT x= 3, what is the value of that minimum? That is, what is f(3)?
     
  4. Dec 11, 2008 #3
    I plugged in 3 for f(x) and got -9 + k = f(x)
     
  5. Dec 11, 2008 #4
    what di I do about the K?
     
  6. Dec 11, 2008 #5
    drand,
    Perhaps the point of the k is that you plug in a (K)onstant that will make the function equal to what you have listed. So if x=3 is the x-coordinate of the minimum, and you are saying it needs to be 11, what value could you put in for k that will make the equation true for both criteria?
     
  7. Dec 11, 2008 #6
    Oh, so because at the min pt is at (3,-9), I just have to shift the graph up so that the min pt would be (3,11). This means that the k value would have to be 20, right?
     
  8. Dec 11, 2008 #7

    HallsofIvy

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    No, the minimum point is NOT at (3, -9). It is at (3, -9+ k) and you want -9+ k= 11. Just solve that equation. (You get the same answer of course.)
     
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