A quick graph of this function makes it look like the roots are probably messy. For these cases, I find that using Newton's method of iterating the root works nicely.

The derivative of the function

F(x) = x^3 - 4x + 2

is

F'(x) = 3x^2 - 4

So the iteration function becomes

x

x

x

If we start with x=-2.2, after 3 iterations we get

The derivative of a*x^b is a*b*x^(b-1). The derivative of a sum is the sum of the derivatives.

The derivative of the function

F(x) = x^3 - 4x + 2

is

F'(x) = 3x^2 - 4

So the iteration function becomes

x

_{new}= x - F(x)/F'(x) = (x*F'(x) - F(x))/F'(x)x

_{new}= (3x^3 - 4x - x^3 + 4x - 2)/(3x^2 - 4)x

_{new}= 2(x^3 - 1)/(3x^2 - 4)If we start with x=-2.2, after 3 iterations we get

**x = -2.21432**If we start with x=0.5, after 3 iterations we get**x = 0.5391889**If we start with x=1.7, after 3 iterations we get**x = 1.6751309**The derivative of a*x^b is a*b*x^(b-1). The derivative of a sum is the sum of the derivatives.