how many turning point are there in this function
y=2x^3-x-1
To find the turning points of the function, we need to find the critical points by taking the first derivative and setting it equal to zero.
The first derivative of y = 2x^3 - x - 1 is:
dy/dx = 6x^2 - 1
Setting this derivative equal to zero:
6x^2 - 1 = 0
Solving for x, we have:
6x^2 = 1
x^2 = 1/6
x = ±√(1/6)
So there are two critical points at x = √(1/6) and x = -√(1/6).
Since the function is a cubic function, it can have at most two turning points. Therefore, there are two turning points in this function.