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Integration and tangents help

  1. Nov 23, 2005 #1
    Hello Everyone,

    I'm trying to do this question:

    Find a function f such that f'(x)=x^3 and the line x+y=0 is tangent to the graph of f.

    Now to find the general equation of f(x) all you have to do is take the extremely basic integral of F'(x) which is going to be f(x) = x^4/4 +C.

    Now the question asks you to find the equation so that f'(x) and x+y=0 is tangent to this equation. I really have no idea how to go about doing thing to be honest. At first I thought that maybe you could set f'x equal to x+y =0 solve, and then set the answer equal to f(x) but I did this and my answer was horrible wrong. Any suggestions on how to go about this question?
     
  2. jcsd
  3. Nov 23, 2005 #2

    Tide

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    HINT: The slope of the line x + y = 0 is -1 and you are to find the point at which this line is tangent to the function whose derivative is given as a function of x.
     
  4. Nov 24, 2005 #3
    Oh that helps. I don't know why I always miss the simple things like that. Thanks, I'll go back now and see if i can figure it out. I appreciate it.
     
  5. Nov 24, 2005 #4
    Ok, so I now know that the function must be as follows:

    f(x)= -x^4/4 +C

    Now just by graphing you can tell that the value of C must also be -1 in order for the line y=-x to be tangent to the graph of f(x), but I cannot seem to show this mathematically.
     
  6. Nov 24, 2005 #5
    Crap, I just noticed that the function cannot equal -x^4/4 it has to be positive so the derivative can be positive, meaning that my answer has to be x^4/4 +1, which I still am unsure of how to show that C must equal 1 mathematically. Any suggestions?
     
  7. Nov 24, 2005 #6

    Tide

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    HINT: At what point does f ' (x) = -1?
     
  8. Nov 24, 2005 #7
    Haha, sometimes I cannot believe myself, ok here is my solution:

    when f'(x) =-1, y=1
    so if you subsitute these x and y values into the equation of the original function and solve for C you get :

    f(x) = x^4/4 +3/4
     
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