Question #c6955

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Answer 1 · 2017-06-09T19:36:31

$"SP": (1/3sqrt 2, 4/9sqrt2), (-1/3sqrt2, -4/9sqrt2)$

$f(x) = 2x-3x^3$

In order to find the stationary points if this function, we need to find $f'$ and solve it for $x$ .

$f'= 2-9x^2$

Let $f' = 0$

Then $2-9x^2 = 0$

$9x^2=2$

$x^2=2/9$

$x= +-sqrt2/3$

We have found the $x$ -coordinates of the stationary points and we know need to find the $y$ -coordinates.

$f(sqrt2/3) = 4/9sqrt2$

$f(-sqrt2/3) = -4/9sqrt2$

So the SP are $(1"/"3sqrt2, 4"/"9sqrt2) and (-1"/"sqrt2, -4"/"9sqrt2)$ .