How Do You Find a Function Whose Integral is x^3 and Has a Tangent Line x+y=0?

[flyers]

Homework Statement

Find the equation f(x) who's integral is \int x^{3} and has a tangent x+y=0

Homework Equations

The Attempt at a Solution

I know that f(x) is 1/4x⁴+c because of the integral. The tangent is the derivative of f(x) at some point

i have the equations

y=x³+c
y=-x

but solving these equations gives me two unknowns...

 

[Char. Limit]

Well, at what point does x^3 equal negative 1 (the slope of y=-x)

 

[Mark44]

What you have written is not very clear. From what I can tell, f(x) = (1/4)x⁴ + C.

You're given that f is tangent to the graph y = -x. This means that f'(x) = x³ has to equal -1 (the slope of the line y = -x). It also means that the graph of f has to have a point in common with the line y = -x.

 

[flyers]

-1?
so

1/4x⁴+c=1
1/4(-1)+c=1
c=3/4

f(x)= 1/4x⁴+3/4

 

[Mark44]

-1?

-1 for what?

so

Why 1? You're not explaining what you're doing, which makes it extremely difficult to understand your work.

1/4x⁴+c=1
1/4(-1)+c=1
c=3/4

f(x)= 1/4x⁴+3/4

 

[flyers]

Sorry, I was replying to Char. limit's question

 

[Mark44]

Can you post the problem exactly as it is worded? I'm having a hard time believing this is what you have actually been given:

Find the equation f(x) who's integral is \int x^{3}
and has a tangent x+y=0

especially the part that says "who's integral is \int x^{3}..."