# determmine image of $f$

Let $f(x)=\dfrac{1}{x+1}\sqrt{x^2-1}$

$D_f={(-\infty,-1)}\cup {[1, +\infty)}$

How can i show algebraically that $$f\left( (1,+\infty) \right)=(0,1)$$

Indeed,

let $y\in \mathbb{R}$ \begin{align} y\in f\left( ]1,+\infty[ \right) &\iff f(x)=y\\ &\iff \dfrac{1}{x+1}\sqrt{x^2-1}=y\\ &\iff \sqrt{x^2-1}=yx+y\\ &\iff \begin{cases} |x^{2}-1|=y^{2}(x+1)^{2}& \\ yx+y>0& \\ x> 1 &\\ \end{cases} \\ &\iff \begin{cases} x-1=y^{2}(x+1)& \\ y(x+1)>0& \\ x> 1 & \end{cases}\\ &\iff \begin{cases} x-y^{2}x=y^{2}+1& \\ y(x+1)>0& \\ x> 1 & \end{cases}\\ &\iff \begin{cases} x=\dfrac{y^{2}+1}{1-y^{2}}& \\ y(x+1)>0& \\ x> 1 & \end{cases}\\ &\iff \dfrac{y^{2}+1}{1-y^{2}}>1 \end{align}

i'm trying to show that $y\in (0,1)$

I'm stuck here

• For a general function, it may be very difficult to determine the image. This one presents relatively few problems, though. Nov 30, 2016 at 18:26
• we can use $h(x)=(f(x))^{2}$ such that $h(x)=\dfrac{x-1}{x+1}$ Nov 30, 2016 at 19:19

for $x>1$

$$f(x)=\sqrt{1-\frac{2}{x+1}}$$

we see that $f$ is continuous and strictly increasing at $(1,+\infty)$.

but

$$\lim_{x\to 1^+}f(x)=0$$

and

$$\lim_{x\to+\infty}f(x)=1$$

thus

$$f((1,+\infty))=(0,1)$$

algebraically

let $x>1$

$$f(x)=y\implies 1-\frac{2}{1+x}=y^2$$ $$\implies x=\frac{2}{1-y^2}-1$$

• $$x>1\implies 0 <\frac{2}{1+x}<1\implies 0<y<1$$

• $$0<y<1\iff 0<1-y^2<1\implies \frac{1}{1-y^2}>1\implies x>1$$

• this is Analytical Approach i'm looking for algebric one Nov 30, 2016 at 15:41
• @Yacob Please look now. Nov 30, 2016 at 16:14
• how did you get that $f(x)=y\iff x=\frac{2}{1-y^2}-1$ please elaborate your steps Nov 30, 2016 at 16:25
• @Yacob And now, You see. Nov 30, 2016 at 16:32
• yes i see thank you Dec 1, 2016 at 20:26