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How come this factoring works for finding points on a curve that equals slope 1?

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]y\prime=4x^3+6x^2+2x+1[/tex]



    2. Relevant equations



    3. The attempt at a solution
    [tex]4x^3+6x^2+2x+1[/tex]

    [tex]-1 = 2x(2x^2+3x+1)[/tex]

    (I moved the +1 over so I don't get 1/x, but I figured this should screw up the equation since it is no longer 0 on the left hand side.)

    [tex]-1 = 2x(2x+1)(x+1)[/tex]

    So the "zeroes" are -1/2, -1, and 0, which is actually the correct answers, but why does this factoring still work? Doesn't the left hand side have to be 0? It bothers me when I can get the correct answer but I don't know why... Isn't that why they're called "zeroes"?

    They original equation is [itex]y = x^4 + 2x^3 + x^2 + x + 1[/itex], where deriving gets me the equation that is stated above. Usually pllugging in an x will solve for slope, but it is already given that the slope is 1. I also used common sense to figure out if x was 0, only the constant 1 would remain and 1 = 1, so 0 is a point for sure.
     
  2. jcsd
  3. Oct 29, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    They are the "correct answers" to what question? They are NOT "zeroes" of the original equation nor do they satisfy that equation.

    So the original problem was to find where [itex]y'= 4x^3+ 6x^2+ 2x+ 1= 1[/itex]? If you subtract 1 from both sides of that, you get [itex]4x^3+ 6x^2+ 2x= 2x(2x+1)(x+1)= 0[/itex].

    I don't know why you were setting [itex]4x^3+ 6x^2+ 2x+ 1[/itex] equal to 0, that makes no sense at all.

    Looks to me like you made two mistakes that just happened to cancel each other!
     
    Last edited: Oct 29, 2009
  4. Oct 29, 2009 #3

    Mark44

    Staff: Mentor

    As HallsofIvy points out, you didn't provide the complete question. Based on what you report as the correct answers and some reverse engineering, the question might have been this:
    "If y' = 4x3 + 3x2 + x, find all values of x for which y' = 0."
    Note that there is no constant term.

    Solution:
    y' = 0
    ==> 4x3 + 3x2 + x = 0
    ==> x(4x2 + 3x + 1) = 0
    ==> x(2x + 1)(x + 1) = 0
    ==> x = 0, or x = -1/2 or x = -1
     
  5. Oct 29, 2009 #4
    The original question was probably: find the points on the curve of x^4+2x^3+x^2+x+c where the slope is equal to 1. After differentiation you get

    [tex]
    y\prime=4x^3+6x^2+2x+1 = 1
    [/tex]

    after canceling the 1's you get an equation you can factor.
     
  6. Oct 29, 2009 #5

    Mark44

    Staff: Mentor

    The point that HallsOfIvy, willem2, and I are making, is that we shouldn't have to guess what the problem is. You need to provide the problem, exactly as stated.
     
  7. Oct 29, 2009 #6
    Sorry. The original question is:

    "Find the equation of the tangent and normal lines to the curve

    [tex]y=x^4+2x^3+x^2+x+1[/tex]

    at the point where the tangent line has slope equal to 1.

    So I derived

    [tex]
    y\prime=4x^3+6x^2+2x+1
    [/tex]

    and went from there.
     
  8. Oct 29, 2009 #7

    Oh! Sorry didn't read this earlier, but you're right. That's where the 1 goes. I forgot to plug in the slope of 1 which cancels the 1 on the other side. I was wondering why it worked even though there was a -1 on the other side.

    Thanks a lot! Appreciate it :D
     
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