Matematické Fórum

stuart clark · 04. 11. 2014 03:05

Maximum value of $f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,$ Where $x>1$

Pavel · 17. 01. 2015 11:02

↑ stuart clark:

$\mathrm{Dom}_f=\mathbb{R}\setminus\{\sqrt[3]{-1\pm\sqrt 2}\}$ . Both numbers where the function f is not defined are less than 1, so f is a continous function for $x>1$ . Let us find the derivative of f:

$f'(x)=\frac{-2x(x^8-2x^6-x^5-x^3+2x^2-1)}{(x^6+2x^3-1)^2}$

Let us denote $P(x)=x^8-2x^6-x^5-x^3+2x^2-1$ . Then the complex numbers $\mathrm i$ and $-\mathrm i$ are roots of $P(x)$ , therefore $x^2+1$ divides $P(x)$ and

$P(x)=(x^2+1)\cdot Q(x),\quad Q(x)=x^6-3x^4-x^3+3x^2-1.$

The equation $Q(x)=0$ is an algebraic reciprocal equation. Let us divide it by $x^3$ and modify it in the following way:

$Q(x)&=0\\ x^3-3x-1+\frac 3x-\frac 1{x^3}&=0\\ \left(x^3-3x+\frac 3x-\frac 1{x^3}\right)-1&=0\\ \left(x-\frac 1x\right)^3-1&=0$

The equation has only two real roots: $x_1=\frac 12(1+\sqrt 5)$ and $x_2=\frac 12(1-\sqrt 5)$ . Therefore $Q(x)$ is divisible by $x^2-x-1$ and

$Q(x)=(x^2-x-1)\cdot R(x),\quad R(x)=x^4+x^3-x^2-x+1,$

moreover $R(x)$ has no real roots and $R(x)>0$ for $x>1$ .

Using the previous results we can express the derivative f' in the following way:

$f'(x)=\frac{-2x(x^2+1)(x^2-x-1)R(x)}{(x^6+2x^3-1)^2}$

If $x>1$ then there is only one stationary point, $\color{blue}{x_0=\frac 12(1+\sqrt 5)}$ . Since $f'(x)$ is positive on $(1,x_0)$ and negative on $(x_0,+\infty)$ , the point $x_0$ is a global maximum of $f$ on $(1,+\infty)$ . Finally,

$\color{blue}\max_{x\in(1,+\infty)}\frac{x^4-x^2}{x^6+2x^3-1}=f(x_0)=\frac 16.$

stuart clark · 22. 01. 2015 09:28

↑ Pavel: Thanku pavel for Nice explanation.

My Solution ::

Given $\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,$ Where $x>1$

So We can Simplify $\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1} = \frac{x^3\cdot \left(x-\frac{1}{x}\right)}{x^3\cdot \left(x^3-\frac{1}{x^3}\right)+2} = \frac{\left(x-\frac{1}{x}\right)}{\left(x^3-\frac{1}{x^3}\right)+2}$

Now Let $u = \left(x-\frac{1}{x}\right)>0\;,\left(x^3-\frac{1}{x^3}\right) = \left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\;,x>1.$

So $\displaystyle f(u) = \frac{u}{u^3+3u+2}=\frac{u}{u^3+u+u+u+1+1}\leq \frac{u}{6\sqrt[6]{u^3\cdot u \cdot \cdot u \cdot u\cdot 1\cdot 1}} = \frac{1}{6}$

Using $\bf{A.M\geq G.M}$ and above equality hold when $\displaystyle u = 1\Rightarrow \left(x-\frac{1}{x}\right) = 1\Rightarrow x= \frac{\sqrt{5}+1}{2}$

Pavel · 22. 01. 2015 09:57

↑ stuart clark:

Nice solution. I suspected that mean inequalities were an appropriate tool to solve the problem. However, I did not find the right modification of the fraction.

Matematické Fórum

#1 04. 11. 2014 03:05

maximum value

#2 17. 01. 2015 11:02

Re: maximum value

#3 22. 01. 2015 09:28

Re: maximum value

#4 22. 01. 2015 09:57

Re: maximum value

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