Idea to find range of a function

Question

I have to find range of this function $$ f(x) = \frac{x}{x^2+x+1} + \frac{x^2}{x^4+x^2+1} $$

I did it by using derivation as follows: $D_f(x)=\mathbb{R}$ so it suffices to check $\lim_{x \to \pm\infty } f(x)$ and $\max/\min$ points.

So I find $$ f'(x) = \frac{x(x+2)}{(x^2+x+1)^2} - \frac{2x(x^4-1)}{(x^4+x^2+1)^2} $$ so $$ f'(x) = -\frac{x^6+2x^4-2x^2-1}{(x^2-x+1)^2 (x^2+x+1)^2} = 0 \quad\implies\quad x = \pm1 $$ Finally, $$ \begin{align} \lim_{x \to -\infty} f(x) &= 0 \\ \lim_{x \to \infty} f(x) &= 0 \\ f(1) &= \frac23 \\ f(-1) &= -\frac{2}{3} \end{align} $$ so $$ R_f = \biggl[ -\frac{2}{3}, +\frac{2}{3} \biggr] $$ Now my question is:

Is there another way to find range, like substitution or ...

Thanks for any hint.
Remark: I tried $x = \tan a$ (years ago) to obtain the range of $$ y = \frac x{x^2+1} \quad\implies\quad y = \frac 12 \sin (2a) $$ but it seems this substitution doesn't work here.

Answer 1

Noticing that if $x\ne \pm 1$,$$ \begin{aligned} f(x) & =\frac{x}{x^2+x+1}+\frac{x^2}{x^4+x^2+1} \\ & =\frac{x(x-1)}{x^3-1}+\frac{x^2\left(x^2-1\right)}{x^6-1} \\ & =\frac{x^5-x}{x^6-1} \\ & =\frac{x\left(x^4-1\right)}{x^6-1} \end{aligned} $$ therefore for any $x\ne \pm 1,f(-x)=-f(x)$ and $f(-1)=-\frac 23=-f(1)$ , $f(-(-1))=\frac 23 =-f(-1)$. Hence $f$ is an odd function.

Now let $g(x)=\dfrac{x}{x^2+x+1}$, then $$ f(x) =\frac{x}{x^2+x+1}+\frac{x^2}{x^4+x^2+1} =g(x)+g\left(x^2\right) $$ We are going to investigate the range of $g(x)$ by letting $$ \begin{aligned} & \qquad \quad y=\frac{x}{x^2+x+1}, \quad \forall x \in \mathbb{R} \\ & \Rightarrow \quad y x^2+(y-1) x+y=0 \quad \forall x \in \mathbb{R} \\ & \Rightarrow \quad \Delta \geqslant 0 \Rightarrow(y-1)^2-4 y^2 \geqslant 0 \\ & \Rightarrow \quad 3 y^2+2 y-1 \leqslant 0 \\ & \Rightarrow \quad(3 y-1)(y+1) \leqslant 0 \\ & \Rightarrow \quad-1 \leq y \leq \frac{1}{3} \end{aligned} $$ $$g_{max}=\frac 13 \textrm{ at }x=1 \textrm{ and } g_{min}=-1 \textrm{ at }x=-1 $$ As $f(x)$ is odd, therefore we only need to study the extrema of $f(x)$ for $x\ge 0.$ For $x\ge 0$, both $g(x)$ and $g(x^2)$ attains their minimum at $x=0$ and maximum values $\frac 13$ at respectively $x=1$ and $x^2=1$. For $x>0$, we can conclude that $f_{max}= \frac 13+ \frac 13= \frac 23$ at $x=1$. As $f(x)$ is odd, therefore the minimum value of $f(x)$ is $-\frac 23.$ Hence the range of $f(x)$ is $[-\frac 23, \frac 23].$

Answer 2

Here's a crack at the problem without any calculus. But it's a bit complex.

1. Show that $f(x) = f(1/x),\quad x \ne 0$. Thus, we need only consider $x \in [-1,1]$.
2. Show that $f$ is odd. Thus, the min of $f$ is equal to negative the max of $f$. So, we now only consider $x \in [0,1]$.
3. For $x\geq 0$, $f(x) = g(x) + g(x^2) \leq 2\max\{g(x), g(x^2)\}$.
4. Show that $g(x) = x/(1+x+x^2)$ is injective on $[0,1]$ (so that it is either increasing or decreasing on the interval).
5. Conclude that only when $x = 0$ or $x = 1$ will $g$ (and $f$) take on its max value.
6. Check that max occurs at $x = 1$.

Answer 3

First, we have $$f(x)-a=-\frac{ax^4-x^3+ax^2-x+a}{(x^2-x+1)(x^2+x+1)}$$

For $a=0$, the numerator of $f(x)-a$ has leading term $x^3$, so $f(x)-a$ changes sign and $0$ is not an extremum.

Since $f(x)\to0$ for $x\to\pm\infty$, the extrema of $f$ are not at infinity, and they are reached at finite values of $x$.

Therefore at extrema $a$ (either min or max), we must have $f(x)-a=0$.

The denominator of $f(x)-a$ is positive, so we are going to consider the numerator. It's a polynomial of degree $4$ unless $a=0$, so it cuts the $x$ axis an even number of times, counting multiplicity. We want $f(x)-a$ to be either always nonpositive or always nonnegative, so it must not cross the $x$ axis. However, the $x$ axis is reached. That is, at extrema $f(x)-a$ has a multiple root.

Let $P=ax^4-x^3+ax^2-x+a$. Any multiple root of a polynomial $P$ is also a root of $P'$. So it's also a root of the GCD.

By the usual (tedious...) euclidean algorithm, we find, with $U=24a(1-a^2)x^2+(38a^2-18)x+a(36a^2-11)$ and $V=6a(a^2-1)x^3+(6-11a^2)x^2+a(1-6a^2)x+4-4a^2$,

$$UP+VP'=36a^4-7a^2-4=(3a-2)(3a+2)(4a^2+1)$$

This has to be zero when $x$ is the abscissa of either extremum, and the last factor is always positive, therefore the extrema are $\pm\frac23$.

Answer 4

After a day I have an Idea. let me show you my work.
$$f(x)=\frac{x^2}{x^4+x^2+1}+\frac{x}{x^2+x+1}\\f(x)=\frac{x^2}{x^4+x^2+1}\frac{\div x^2}{\div x^2}+\frac{x}{x^2+x+1}\frac{\div x}{\div x}\\f(x)=\frac{1}{x^2+1+\frac{1}{x^2}}+\frac{1}{x+1+\frac{1}{x}}$$ by using $a=x+\frac 1x \to a^2=x^2+\frac{1}{x^2}+2$ and $|a|\ge 2$ we have
$$y=\frac{1}{a^2-1}+\frac{1}{a+1}\\y=\frac{1+a-1}{a^2-1}\\a^2y-y=a \\a=\frac{1\pm \sqrt{1+4y^2}}{2y}$$so now $$x>0 \to a\ge2\to 1+ \sqrt{1+4y^2}\ge 4y \\\to 1+4y^2\ge (4y-1)^2 \\12y^2-8y \le 0 \\4y(3y-2)\le 0 \to 0\le y\leq \frac32$$ also for $x<0 \to a\le -2 \to -\frac32 \le y \le 0 $ finally $$-\frac32 \le y\le \frac 32$$