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Homework Help: Calc Question

  1. Mar 20, 2006 #1
    Let f be the function given by f(x)= (x^3)-(6x^2)+p, where p is an arbitrary constant.

    a)Write an expression for f'(x) and use it to find the relative maximum and minimum values of f in terms of p. Show the analysis that leads to your conclusion.

    b)For what values of the constant p does f have 3 distinct roots?

    c)Find the value of p such that the average value of f over the closed interval [-1,2] is 1.

    For a), I found the derivative, which is f'(x)=(3x^2)-12x. I set that equal to 0, which gives me x=0,4. I put these values back into the original equation and got f(0)=o and f(4)=-32. In terms of p, I got 0=p and 32=p. Is this right?

    I have no clue how to do b). I kind of know how to do c) but I would appreciate some help with it. Thanks a lot.
  2. jcsd
  3. Mar 21, 2006 #2
    a) Your maximum and minimum values of f (f(0) and f(4)) should be in terms of p, as specified in the problem. p is a constant that is the same for any given case of the function. You have no requirement that f(x)=0, so you can choose whatever values of p you desire.

    b) Try graphing f with p=0, and consider how changing the value of p would affect the graph.
  4. Mar 22, 2006 #3
    For a), I've got the points in terms of p.

    For b), I'm just not quite sure what they are asking for when they say "distinct roots."

    For c), I know that to find the average value, you take 1/(-1-2) and multiply that by the definite integral of f'(x) from -1 to 2. Since it has to equal 1, do 1=(-1/3)(?) I am not sure what to put where the question mark is.
  5. Mar 22, 2006 #4


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    No, you don't. For some reason, you calculated f(0) and f(4), assuming p was 0, and then set that equal to p???
    f(x)= x3- 6x2+ p. What is f(0)? What is f(4)? Do you see that the answer to those question still has "p" in it?

    "roots"- solutions to the equation. "distinct"- different.
    A cubic equation such as x3- 6x2+ p= 0 may have 3 'distinct' solutions or it may have a "double root" or a triple root. What values of p guarentee that this equation does not have a double or triple root? (Hint: if a, b, c are distinct roots of polynomial Q(x) then Q(x)= d(x-a)(x-b)(x-c) so Q'(x)= d(x-b)(x-c)+ d(x-a)(x-c)+ d(x-a)(x-b). Setting x= a, b, c does NOT make that 0. If Q has a as double root and other root b then Q(x)= (x-a)[suQ]2[/suQ](x-b) so Q'(x)= 2(x-a)(x-b)+ (x-a)[suQ]2[/suQ] and Q'(a)= 0. If Q has a as triQle root, then Q(x)= (x-a)[suQ]3[/suQ] so Q'(x)= 3(x-a)[suQ]2[/suQ] and again Q'(a)= 0.) What values of Q guarentee that the same x value cannot make both Q and Q' equal to 0 (what values of x make Q'= 0? What is the value of Q(x) for those values of x?)

    Okay, so go ahead and do the integral! What is
    [tex]\int_{-1}^2 (x^3- 6x^2+ p)dx[/tex]
    THAT'S what you want.
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