Calculus problem

Question

find a function such that f'(s)=5x^3 and the line x+y=0 is tangent to the graph of f

Answer 1

As f'(x) = 5 x^3, f(x) = 5/4 x^4 + C

x+y = 0 has slope -1 and is tangent at (x, f(x)) = (x,-x) (this is the coordinate of points on x + y = 0

f'(x) = 5x^3 = -1 at x = (-1/5)^(1/3)

Then, y = -x = (1/5)^(1/3) at this point

Thus, f(x) =5/4 x^4 + C = (1/5)^(1/3)

5/4((-1/5)^(1/3))^4 + C = (1/5)^(1/3)

C = (1/5)^(1/3) - 5/4((-1/5)^(1/3))^4 =

(1/5)^(1/3) - 5/4 * 1/5 * (1/5)^(1/3) =

3/4 * (1/5)^(1/3)

f(x) = 5/4 x^4 + 3/4 * (1/5)^(1/3) or 5/4 x^4 + 3/(4 * 5^(1/3))