Find a function f such that f '(x) = 3x3 and the line 3x + y = 0 is tangent to the graph of f.

Question

f(x) =

Answer 1

dy/dx = 3 x^3
then
y = (3/4) x^4 + c

slope of line y = -3x + 0 = -3
so where does our slope = -3?
3 x^3 = -3
x = 1
so make y the same for the function and the line at x = 1
line at x = 1 is y = -3
so
(3/4)x^4 + c = -3 at x = 1
3/4 + c = -3 = -12/4
c = -15/4
so
f(x) = (3/4) x^4 -15/4
or
4 y = 3 x^4 - 15