1. ## stationary points

I got a question to find stationary points of: y=(x^3) - 3x

now i differentiated it and set to 0 and get

3x^2 - 3 = 0 now am not sure from here what to do?

2. Originally Posted by taurus
I got a question to find stationary points of: y=(x^3) - 3x

now i differentiated it and set to 0 and get

3x^2 - 3 = 0 now am not sure from here what to do?
solve for x of course. and then find the y coordinate for the corresponding x-values

3. i did that and get 1,-2 which i dont think is right

4. Originally Posted by taurus
i did that and get 1,-2 which i dont think is right
The solutions of:

$3x^2-3=0$

are $x=\pm 1$

RonL

5. when i saw the answers they were:

(-3/2, -5/4)

?

6. Originally Posted by taurus
when i saw the answers they were:

(-3/2, -5/4)

?
If these are supposed to be the x values for which the slope of x^3-3x is
zero then they are wrong.

RonL

7. the question is asking for the stationary points for the function:
y = x^3 - 3x

8. Originally Posted by taurus
the question is asking for the stationary points for the function:
y = x^3 - 3x