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).
Best Answer
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.