Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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).

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.