Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 67

The tangent line must touch a point on the f(x) function.
Set both equations equal to each other.
kx^3 = x+1
Take the derivative of f(x) to find another relationship with k and x. The k is a constant, so do not eliminate it!
f(x)= kx^3
f'(x)= 3kx^2
Set this equal to the slope of the tangent line given, 1.
1= 3kx^2
Solve for k, since this is easier than solving for x.
k = 1/(3x^2)
Substitute the k back into the first equation.
(1/(3x^2))x^3 = x+1
x/3 = x+1
Multiply three on both sides.
x= 3x+3
-2x =3
x= -3/2
Re-substitute the x back to the first equation to solve for k.
k(-3/2)^3 = -3/2+1
k(-27/8) = -1/2
k=1/2 * 8/27 = 4/27