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Find the Co-ordinates of the stationary points on the curve $$f(x)=2x^3-4x^2+2$$

What I did was to differentiate $f(x)$ then factorise to find to possible $x$ values then put those two values into $f(x)$.

Although I came out with $$f(0) = 2,$$ $$f(4/3) = 13.85185$$

Where have I gone wrong? Thanks

jackdh
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  • You have a different equation in the title and text. Please correct this. –  Apr 19 '13 at 11:22
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    Done sorry, i edited the questions title, although i forgot i also needed to edit the text. – jackdh Apr 19 '13 at 11:23
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    Your method is correct, if you are setting the derivative equal to $0$ which I think you are. Just check the sign of the second value you obtain. –  Apr 19 '13 at 11:26
  • I double check that it would be positive, Are you saying it should be negative or just make sure? – jackdh Apr 19 '13 at 11:31
  • EDIT: I mean the second value of $x$ you obtain should be negative. –  Apr 19 '13 at 11:36
  • It is i missed a - sign thank you – jackdh Apr 19 '13 at 11:40
  • Yes with the edited equation you have the correct $x$ co-ordinates. However, I don't agree with the value of $f(4/3)$. –  Apr 19 '13 at 11:44
  • Also please see http://meta.math.stackexchange.com/q/9017/67881 in concern of typos in questions after answers have been posted. –  Apr 19 '13 at 11:59

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Your method is correct. You should obtain: $$ f^\prime(x) = 0\quad\iff\quad 2x(3x+4) = 0 $$ which is true for $x=0$ and $x=-4/3$. As Stephen pointed out, I think you made a mistake in the sign of your second value. The associated ordinates correspond to $f(x)$ for both values.

Jonathan H
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