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Let $a,\lambda \in \mathbb{R}, \lambda>0$ $$f(x)=\frac{x+\lambda}{x+1}$$ $$a=\frac{a+\lambda}{a+1}\Rightarrow a^2=\lambda\Rightarrow a=\pm\sqrt\lambda$$ $$f'(x)=\frac{1-\lambda}{(x+1)^2}$$

I want to know when $|f'(a)|<1 $, so first I calculate $|f'(a)| $ : $$|f'(a)|=\left|\frac{1-\lambda}{(a+1)^2}\right|=\left|\frac{1-a^2}{(a+1)^2}\right|=\left|\frac{1-a}{1+a}\right|$$

Then I solved the inequality $|f'(a)|<1$: $$\left|\frac{1-a}{1+a}\right|<1\Rightarrow -1<\frac{1-a}{1+a}<1\Rightarrow$$ $$0<\frac{2}{1+a}\,\hspace{0.25cm} \text{and}\hspace{0.25cm} \frac{2a}{1+a}>0 $$ $$-1<a\,\hspace{0.25cm} \text{and}\hspace{0.25cm} (-1>a \hspace{0.25 cm} \text{or} \hspace{0.25cm}a>0) $$ $$\boldsymbol{a>0} $$ So $a$ should be $a=\sqrt\lambda$ to $|f'(a)|<1$.

Is my work correct? or Is there a nother way to do it?

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  • $\begingroup$ For what variable want you to solve this inequality? $\endgroup$ – Dr. Sonnhard Graubner Apr 23 '18 at 16:09
  • $\begingroup$ @Dr.SonnhardGraubner for $a$ $\endgroup$ – B. David Apr 23 '18 at 16:11
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Hint: you can write $$\sqrt{|1-\lambda|}\le |a+1|$$ So you have two cases: $$a\geq -1$$ then you will get $$\sqrt{|1-\lambda|}-1\le a$$ And if $$a<-1$$ then you will get $$\sqrt{|1-\lambda|}\le -a-1$$ or $$a\le -\sqrt{|1-\lambda|}-1$$

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  • $\begingroup$ So my work is wrong?, It doesn't matter the relation between $\lambda$ and $a$ that I wrote previously to the inequality? $\endgroup$ – B. David Apr 23 '18 at 16:28
  • $\begingroup$ You have asked for an other way, and this is my way. $\endgroup$ – Dr. Sonnhard Graubner Apr 23 '18 at 16:31

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