how many turn around points are in the function f(x)=(x-1/3)^2(x-2)^3

f(x) has two roots

it touches the x-axis at x = 1/3, so that's one turning point.
It turns again so it can cross the x-axis at x=2

so, 2 turning points

To determine the number of turning points in the function f(x) = (x - 1/3)^2(x - 2)^3, we need to examine the behavior of the function and identify the points where it changes direction. A turning point occurs when the function changes from increasing to decreasing or vice versa.

To find the turning points, we follow these steps:

1. Calculate the derivative of the function f(x) with respect to x. The derivative gives us the slope of the function at any given point.

f'(x) = 2(x - 1/3)(x - 2)^3 + (x - 1/3)^2(3(x - 2)^2)

2. Find the critical points by setting the derivative equal to zero and solving for x.

2(x - 1/3)(x - 2)^3 + (x - 1/3)^2(3(x - 2)^2) = 0

This equation may seem complicated, but we can simplify it further.

3. Expand and simplify the equation:

2(x - 1/3)(x - 2)^3 + 3(x - 1/3)^2(x - 2)^2 = 0

4. Solve the equation for x. This may require factoring or using other algebraic techniques.

After solving for x, you will obtain certain values that correspond to the critical points.

5. Finally, count the number of distinct values of x obtained in step 4. Each distinct value of x corresponds to a turning point in the function.

Counting the number of distinct turning points will give you the answer to the original question.

Note: Depending on the complexity of the equation and the difficulty of solving for x, this process may be more straightforward or more involved.