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Anti Derivative Question

  1. Dec 7, 2007 #1
    Here is the question:

    Find a function f which satisfies both of the following properties:
    f ' (x) = x^3
    The line x + y = 0 is tangent to the graph of f.

    I figured out that f(x) is 1/4x^4 + C. Now I don't know what to do. I know I need to figure out C but i'm stuck. I isolated x+y=0 for y to get y= -x, and the derivative of that is -1, so the slope of the tangent line is -1. So I then figured out what x value causes x^3 to also be -1, and it turns out to be -1. This is the x-coordinate at which the line is tangent to. So the y coordinate would be -1 + y=0. So y is 1. The point at which the line is tangent to f is (-1,1). If everything up to this point is correct, how do I find C?
  2. jcsd
  3. Dec 7, 2007 #2


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    What equation does/should the general solution satisfy at (-1,1)?
  4. Dec 7, 2007 #3
    Do I go 1=1/4(-1)^4 + C and solve for C?
  5. Dec 8, 2007 #4


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    Yes. The important thing is understanding why you can do that. In order that the graph of your function be tangent to x+ y= 0 at x= -1, the graph has to pass through (-1,1). That is true only if 1= (1/4) (-1)^4+ C.
  6. Dec 8, 2007 #5
    Thank you so much for explaining that to me, now I can go write my Math 110 final....yay?
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