# Problem of the week #96 - January 27th, 2014

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#### MarkFL

Staff member
This week's problem was submitted for our use by anemone:

Without using the calculus, determine the minimal point(s) of the following function:

$$\displaystyle f(x)=\frac{2+12x^4}{x^2}$$

Thank you anemone for this problem. --------------------

#### MarkFL

Staff member
Congratulations to the following members for their correct solutions:

1) mente oscura

Honorable mention goes to both kaliprasad and Pranav for finding the correct $y$-coordinate, but a minor error in computing the corresponding $x$-coordinate was made.

Here is mente oscura's solution:

$$y=\dfrac{2+12x^4}{x^2}$$

$$12x^4-yx^2+2=0$$

$$x^2=\dfrac{y \pm \sqrt{y^2-96}}{24}$$

$$For \ y^2<96 \rightarrow{} \ not \ real \ solution \rightarrow{} For \ y^2=96 \ "y" \ is \ minimal$$

$$y=\sqrt{96}=\sqrt{4^26}= 4 \sqrt{6} \rightarrow{} x^2=\dfrac{4 \sqrt{6}}{24}= \dfrac{\sqrt{6}}{6}=\dfrac{1}{\sqrt{6}}$$

$$x= \pm \dfrac{1}{\sqrt{6}}$$

Regards.

Here is the similar solution submitted by anemone:

Set $y=f(x)$ and multiply through by $x^2$ to obtain:

$\displaystyle x^2y=2+12x^4$

Let $u=x^2$ and arrange in standard quadratic form on $u$:

$\displaystyle 12u^2-yu+2=0$

Analysis of the discriminant will reveal the boundary of the range of $y$. We require real roots for $u$, thus the discriminant must be non-negative:

$\displaystyle (-y)^2-4(12)(2)\ge0$

$\displaystyle y^2\ge96$

Since $u=x^2$, we must discard the negative root, and take only the positive root:

$\displaystyle y\ge\sqrt{96}=4\sqrt{6}$

Hence, the range of $y=f(x)$ is:

$\displaystyle \left[4\sqrt{6},\infty \right)$

To determine where this minimum occurs, we may substitute for $y$ and solve for $u$:

$\displaystyle 12u^2-4\sqrt{6}u+2=0$

Divide through by 2:

$\displaystyle 6u^2-2\sqrt{6}u+1=0$

Factor:

$\displaystyle \left(\sqrt{6}u-1 \right)^2=0$

Hence, we find:

$\displaystyle u=x^2=\frac{1}{\sqrt{6}}$

$\displaystyle x=\pm\frac{1}{\sqrt{6}}$

And so we may conclude that the minima for the given function are at:

$\displaystyle \left(\pm\frac{1}{\sqrt{6}},4\sqrt{6} \right)$.

At this time, I would am very pleased to announce that anemone has graciously agreed to take over the posting of our POTW for Secondary School/High School students. I also wish to extend a hearty thanks to Jameson for having done this for 95 weeks straight without missing a beat. So, with no further ado, we now leave this in the very capable hands of anemone. Status
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