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Looking for help with antideratives

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Find a function f such that f'(x)=x^3 and the line x+y=0 is tangent to the graph of f.

    2. Relevant equations

    3. The attempt at a solution
    f(x)=(x^4)/4 + C

    I don't know where to go from here... the tangent line equation is a way to find the slope, no? So rearranging x+y=0 gives y=x which means that the slope is 1....but wait, doesn't y=x mean that the tangent is a straight line, so therefore the derivative is 0? As you can tell, I'm very confused!!
  2. jcsd
  3. Oct 24, 2007 #2
    think more about y=x being a straight line, compare this to something such as y=5, still think its straight? you also have a sign error when rearranging the tangent line equation.
  4. Oct 24, 2007 #3
    Oh I see....thanks =)
    so y=-x
    therefore slope= -1
    I have the equation of the function. Now is the next step just to see what C must equal in order for the slope to be -1?
  5. Oct 24, 2007 #4


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    You have two functions, the straight line f(x)=-x and the curve g(x)=x^4/4+C. At some value of x (let's call it x=a), the line is tangent to the curve.

    What does this tell you about the values of f(a) and g(a)?

    What does it tell you about the values of f'(a) and g'(a)?
  6. Oct 24, 2007 #5
    The value a in f(a) can be substituted into g(a) and the value of a in f'(a) can be substituted into g'(a)??
  7. Oct 24, 2007 #6


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    Ok, you've got the slope is -1. Where is f'(x)=(-1)?
  8. Oct 24, 2007 #7
    so x=-1
  9. Oct 24, 2007 #8


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    Good. Now at the tangent point (x=-1) y=f(x) has to touch the curve y=-x. Can you determine C in your antiderivative?
  10. Oct 24, 2007 #9
    f(x)=(x^4)/4 + C
    f(-1)=1/4 + C
    So since y=-x touches (x^4)/4 + C...and x=-1, then y=1
    1=(1/4) + C

    So f(x)=(x^4)/4 + 3/4 ?
  11. Oct 24, 2007 #10


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    You don't have to ask me if it's right. (It is). If it's slope is -1 and it's value is 1 at x=-1, then you surely are correct. Check it again.
  12. Oct 24, 2007 #11
    Hold on, I'm kind of confused as to why I plug in x=-1 in the antiderivative. I found x=-1 in the derivative form, so why am I plugging it into the antiderivative? What am I looking for there?
  13. Oct 24, 2007 #12


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    The antiderivative is your function y=f(x). The derivative f'(x) is the slope of your function. At the tangent point to a line f(x) should equal the value of the line and f'(x) should equal the slope of the line.
  14. Oct 24, 2007 #13
    Essentially the solution is composed of knowing two things about tangents to curves.

    1) The tangent line and the curve share a single common point.
    2) The slope of the tangent line and the curve are equivalent at that point.

    So after obtaining the antiderivative with some constant C, you seek to determine what value of C you need. Well, you can solve for what value of x the slopes are the same at, and you know what the tangent line equation is, so you can solve y. Looking at (1) and (2) then, we've found the point at which (2) is true for the line and the curve.

    Now all that remained for you to do was plug in that y and x value into the antiderivative, since the antiderivative and the tangent line have to share this common point (see (1)). You then found your value of C, and thus found the curve that fulfilled the above two conditions, and thus solved the problem.
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