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
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.