Dividing numerator and denominator by $x^2$:-

$$\frac{1}{x^2+ \frac{1}{x^2}}$$

Applying AM GM, we get range as $[0, 1/2]$.

But if we are dividing by $x^2$, then the function should not attain value $0$ isn't it? Then is this method wrong?

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Mar 27, 2016 at 12:12
  • $\begingroup$ Just apply derivatives to get maxima,minima $\endgroup$ Mar 27, 2016 at 12:15
  • 1
    $\begingroup$ Just another way : Can you put $$ x^2=\tan y$$ $\endgroup$ Mar 27, 2016 at 12:15
  • $\begingroup$ @labbhattacharjee, good method! $\endgroup$ Mar 27, 2016 at 13:19

2 Answers 2


It is not wrong if you use a little bit of care in your language. Say: for $x = 0$ we get the value $0$. For $x \ne 0$ ... then the rest of your argument; we get the range $(0,\frac 1 2]$. Now combine.


You can do as follows, too:

Clearly the range is contained in the non-negative numbers, so pick $\;c\in\Bbb R_+\;$ :

$$\frac{x^2}{1+x^4}=c\iff cx^4-x^2+c=0$$

For the above biquadratic to have a real solution it must be that the discriminant is non-negative:

$$\Delta=1-4c^2\ge0\iff c^2\le\frac14\stackrel{\text{we know}\;c\ge0}\iff 0\le c\le\frac12$$


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