Range of f(x)

[man111]

The function [tex]f(x)=\sqrt{9x^2+173x+900}-\sqrt{9x^2+77x+900}[/tex] and [tex]f(x)\in (0,M)[/tex] and [tex]x\in(0,\infty)[/tex],then find [tex]M[/tex]

[martin123456]

So [tex]\sqrt{9x^2+173x+900}-\sqrt{9x^2+77x+900}<M \Rightarrow x^2(96^2-36M^2)-x.500M^2+M^4-3600M^2 <0[/tex][tex]\forall x >0[/tex].
If [tex]M<16[/tex] then the leading coeeficient is positive, the product of the roots is negative anf thus there are two real roots, one of which is positivea and so it's impossible the inequality to holds for every x >0.
If [tex]M = 16[/tex]. Then [tex]16^2(-500x+16^2-3600) < 0[/tex] for all positive x. So it holds.
Since [tex]f(x)[/tex] is continuos the above is enough to coclude that that value we search is 16.

[man111]

thanks martin