how many turning point are there in this function

Question

y=2x^3-x-1

Answer 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.