2
$\begingroup$

I just got the set and Ι tried to find the supremum and infimum and prove it. $$(x-2)\sqrt{\frac{x+1}{x-1}} \quad \text{ for } \quad 2< x\leq 54$$

I succeed to get to this set $\frac{x-2}{x-1}\sqrt{x^{2}-1}$ but I'm stuck.

What can I do now ? Thanks.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

Hint: Let $f(x)=\frac{x-2}{x-1}\sqrt{x^2-1}$ and calculate the derivative of $f$ to check that this function is (strictly) increasing for $2<x\le 54$. Moreover, as $x \to 2^+$ this function is continuous.

$\endgroup$
1
  • $\begingroup$ How can i find its increasing without calculate the derivative of $f$ ? $\endgroup$
    – NM2
    Dec 15, 2015 at 10:52
0
$\begingroup$

On one hand, since $x>2$ we have $$ (x-2)\sqrt{\frac{x+1}{x-1}}>0. $$ On the other hand $$ \lim_{x\rightarrow 2^+}(x-2)\sqrt{\frac{x+1}{x-1}}=0. $$ Hence $$ \inf_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=0. $$ Since $2<x\leq 54$ $$ (x-2)\sqrt{\frac{x+1}{x-1}}=\frac{x-2}{x-1}\sqrt{x^2-1}=\left(1-\frac{1}{x-1}\right)\sqrt{x^2-1}\leq \left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}. $$ Hence $$ \sup_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\max_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=\left(1-\frac{1}{54-1}\right)\sqrt{54^2-1}. $$

$\endgroup$
1
  • $\begingroup$ But how do you know its increasing ? I think we have to proove it, No ? $\endgroup$
    – NM2
    Dec 15, 2015 at 10:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .