I am teaching myself the calulus component necessary to get thorugh an econ based stats and applied math class. My algebra is killing me please help - the practive problem is given
$y = –x^3 + 7x – 4 $
Give the relative extrema and points of inflection. (the application is for min and max prices / costs etc given an equation). so I have first and second derivatvies as:
$ y'(x) = -3x^2 + 7 $
$ y''(x) = -6x$
I then set $f'(x) = 0$ and solve
$-3x^2 + 7 = 0$
$-3x^2 = -7$
$x^2 = {-7\over-3} = \frac73$
$x = \sqrt{\frac73} ~= 1.53$
BUT the package tells me the answer is
${1\over\sqrt{21}}$
a couple of questions
1) To find min and max values I solve first derivative for zero AND plug that into the original equation. I htink this is the right approach (Can I get a confirmation)?
2) ARe my derivatives coorect (they look simple enough, and I confirmed them using R (my coding skills much greater than my math skills):
> D(expression(-x^3 + 7*x -4),"x") 7 - 3 * x^2 > D(D(expression(-x^3 + 7*x -4),"x"),"x") -(3 * (2 * x))
3) assuming the above, how does the $-3x^2 + 7 = 0$ get reduced to $x={1\over\sqrt{21}}$ I must be missing some major algebra lesson (it has been 23 years, and while it is not an exuce it is a long time).