How many solutions exist for the given equation x^12 – 4x + 1?

To find how many solutions exist for the equation x¹² – 4x + 1 = 0, we can analyze the polynomial function represented by the left-hand side.

This equation is a polynomial of degree 12. According to the fundamental theorem of algebra, a polynomial of degree n has exactly n roots (solutions), which can be real or complex.

Next, we should consider the behavior of the function f(x) = x¹² – 4x + 1. Since the leading term, x¹², is positive, as x approaches positive or negative infinity, f(x) also approaches positive infinity.

Now, to find the number of real solutions specifically, we can examine the derivative of the function to check for local extrema. The first derivative f'(x) = 12x¹¹ – 4 can help us determine where the function increases, decreases, and whether it has any turning points.

Setting the derivative equal to zero gives us:

12x¹¹ – 4 = 0

From here, we find:

x¹¹ = 1/3

This yields one real solution for x. To understand further, we can evaluate the second derivative for concavity and find whether this point corresponds to a minimum or a maximum.

Since the function is even powered with a positive leading coefficient, it confirms that the function has symmetry and thus means there will be at least two solutions that are complex for every real solution counted.

In summary, while there is one real root, in total, the degree of this polynomial guarantees that there are 12 solutions in total across both real and complex numbers. Therefore, the total number of solutions for the equation x¹² – 4x + 1 = 0 is 12.