# Applied Math for economics question: mostly algebra help

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).

-

Your calculation is largely correct. We have $x^2=7/3$, so $x=\pm\sqrt{7/3}$. It may be that you rejected the negative root because for "physical" reasons the variable $x$ must be non-negative. Or it may be that you forgot about the possibility that $x=-\sqrt{7/3}$.

The number $\dfrac{1}{\sqrt{21}}$ is not a solution of the equation $y'(x)=0$. You can easily verify that the answer you got is a solution by substituting in $y'(x)$, and seeing whether the result is $0$. It is.

There are various alternate ways to write the solutions. For example, we can write the solutions as $\pm\frac{\sqrt{7}}{\sqrt{3}}$. Then you can multiply top and bottom by $\sqrt{3}$ to obtain $\pm \dfrac{\sqrt{21}}{3}$. This procedure is called rationalizing the denominator. (Some people do not like square roots in the denominator.) Perhaps the person solving the problem made a mistake in rationalizing the denominator.

Yes, to find the relative (aka local) maximum and minimum values for your function, you substitute the two values where $y'(x)=0$. To see what's happening, using software or in some other way, graph the function. We hit the top of a hill at $x=\sqrt{7/3}$, and the bottom of a valley at $x=-\sqrt{7/3}$.

Important: Your function does not have a maximum or minimum. For very large positive $x$, it is huge negative, and for very large negative $x$, it is huge positive.

-
I missed the +/- in front of the sqrt. Typo, thanks for catching that. But you are agreeing with me that what the software tells me is correct 1/sqrt(21) is not actually right? That would seem to underscore my inabilty to come up with that solution. – akaphenom Nov 23 '12 at 17:16
Certainly $1/\sqrt{21}$ is not a solution of $y'(x)=0$ for the function $y(x)$ that you wrote down. It is possible that you did not write down the correct $y(x)$. – André Nicolas Nov 23 '12 at 17:23
no chance I copied it wrong, I used cut and paste. I have been finding wrong answers all semested, in areas I have more confidence in I am better at alerting. Since my algebra is poor, I thought i was being dumb. THank you – akaphenom Nov 23 '12 at 17:25
Please see the important comment I added at the end. The function $y(x)$ has neither a maximum nor a minimum. But it does have local (relative) max and min. – André Nicolas Nov 23 '12 at 17:30
It is a good observation, and noted – akaphenom Nov 23 '12 at 17:35