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?
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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: $\displaystyle 3x^2-3=0$ are $\displaystyle x=\pm 1$ 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
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