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How to determine the range of the following function $\frac{x}{1+ |x|}$?

when I calculated it, it was $\mathbb{R}$, but my professor said that the range is ]-1,1[, could anyone explain for me why?

thanks!

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  • $\begingroup$ Why do you think it's $\mathbb{R}$? $\endgroup$ – Dylan Dec 5 '17 at 7:35
  • $\begingroup$ 1) I divided the function into 2 parts to get rid of the absolute value @Dylan $\endgroup$ – user426277 Dec 5 '17 at 7:39
  • $\begingroup$ 2)Then I started with $x \geq 0$ and reached the shape of the function where also the shape of the function was $\geq 0$.@Dylan $\endgroup$ – user426277 Dec 5 '17 at 7:42
  • $\begingroup$ 3) I repeated the same step for $x < 0$reached the shape of the function where also the shape of the function was $< 0$. $\endgroup$ – user426277 Dec 5 '17 at 7:44
  • $\begingroup$ It's true that the function is $\ge 0$ for $x \ge 0$, but does it grow without bound? Or is there a limit as $x \to \infty$? $\endgroup$ – Dylan Dec 5 '17 at 7:45
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Observe that $\large|\frac {x}{1+|x|}|=\frac {|x|}{1+|x|}=1-\frac {1}{1+|x|} \lt 1.$

$\therefore |\frac {x}{1+|x|}| \lt 1 \Rightarrow -1 \lt \frac {x}{1+|x|} \lt 1$.

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  • $\begingroup$ I'd use the strict inequality instead of $\le$. The function can never equal $1$ $\endgroup$ – Dylan Dec 5 '17 at 7:57
  • $\begingroup$ @Dylan Yes yes you are right. $\endgroup$ – Error 404 Dec 5 '17 at 7:59
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For $x\geq 0$, $\dfrac{x}{1+|x|}=\dfrac{x}{1+x}=1-\dfrac{1}{1+x}$, it is increasing on $[0,\infty)$, so it maps onto $[0,1)$. Similarly you can deal with $(-\infty,0]$.

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  • $\begingroup$ You can also get the answer for $x \leq 0$ by using the fact that the function is odd. $\endgroup$ – Joppy Dec 5 '17 at 7:42
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Let $f(x)=\frac{x}{1+ |x|}$. Then: $|f(x)|=\frac{|x|}{1+ |x|} \le 1$, hence

$f( \mathbb R) \subseteq [-1,1]$.

Furthermore: $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.

Show that $f(x) \ne 1$ and $f(x) \ne -1$ for all $x$.

Are you now in a position to derive $f( \mathbb R) =]-1,1[$ ?

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check : $1+|x|\neq 0\Rightarrow |x|\neq-1$, which is true for all $x$.

Now,

when $0\leq x<\infty , y=\frac{x}{1+x}, x<x+1\Rightarrow 0\leq y<1...(I)$

when $-\infty <x<0 , y=\frac{x}{1-x}$ ,

|numerator|<|denominator|, so $|y|$ lies between $0$ and $1$, but since numerator is negative and denominator is positive,$-1<y<0...(II)$

combining $(I),(II)$ we get $-1<y<1$

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