1. ## Calculus Antiderivative Question

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. I don't know if im supposed to do this or how to relate this to C. Please help!

2. Good job ndiggity, you're almost there, and everything you have done so far is on the right track.

All you need to do now is find the value of x for which f'(x) = -1. This must be the point at which x+y=0 is a tangent, as the derivative is the same there. You can then get the y value for that point from x+y = 0 then solve your other equation for C.

3. Ok, so the x value would be -1 then, and therefore the y value is 1. So the point is (-1,1). So do I plug in -1 in the anti-derivative I came up with to find C?

4. ## Exactly right

Ok, so the x value would be -1 then, and therefore the y value is 1. So the point is (-1,1). So do I plug in -1 in the anti-derivative I came up with to find C?
Exactly right