`f(x) = kx^3, y = x + 1` Find k such that the line is tangent to the graph of the function.
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!
Set this equal to the slope of the tangent line given, 1.
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.
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`