Thread: 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?

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

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

Originally Posted by taurus

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

The solutions of:



are

RonL

when i saw the answers they were:

(-3/2, -5/4)

?

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

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

so the answers are wrong?

Originally Posted by taurus

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

so the answers are wrong?

The posibilities are:

1. You have copied the problem incorrectly
2. You have checked the answer to a different question
3. The book is wrong

all of these are common errors

RonL