just checking...

[Sarahisme]

is this a 'proof' for this question?

 

[math]

h(x) = \sqrt(3 + x)

[/math]

 

we want to find a K > 0 such that

[math]

|h(x) - h(y)| \leq K|x-y| \ \ \forall x,y \in [0,1]

[/math]

 

that is we want a K > 0 such that

[math]

|\sqrt(3 + x) - \sqrt(3 + y))| \leq K|x-y| \forall x,y \in [0,1]

[/math]

 

now

[math]

|\sqrt(3 + x) - \sqrt(3 + y)| = \frac{|(\sqrt(3 + x) - \sqrt(3 + y))(\sqrt(3 + x) + \sqrt(3 + y)|}{|\sqrt(3 + x) + \sqrt(3 + y)|}

[/math]

[math]

=\frac{|(3 + x) - (3+y)|}{|\sqrt(3 + x) + \sqrt(3 + y)|}=\frac{|x-y|}{|\sqrt(3 + x) + \sqrt(3 + y)|}

[/math]

 

So, we want:

[math]

\frac{|x-y|}{|\sqrt(3 + x) + \sqrt(3 + y)|} \leq K|x-y|

[/math]

 

The max. value that [math]\frac{|x-y|}{|\sqrt(3 + x) + \sqrt(3 + y)|}[/math] attains is when

[math]

x=y=0

[/math]

 

So when x=y=0

[math]

\frac{|x-y|}{|\sqrt(3 + x) + \sqrt(3 + y)|} = \frac{1}{\sqrt(3)+\sqrt(3)}

= \frac{1}{2\sqrt(3)}[/math]

 

So Let K = [math] \frac{1}{2 \sqrt(3)} [/math]

 

[math]

\therefore \ if \ \ K = \frac{1}{2\sqrt(3)}

[/math]

[math]

then |\sqrt(3 + x) - \sqrt(3 + y))| \leq K|x-y| \ \ \forall x,y \in [0,1]

[/math]