One-sided differentiation proof

[transgalactic]

i am given two differentiable function f and g .

prove that for u(x)=max(f(x),g(x))

and v(x)=min(f(x),g(x))

 

there is one sided derivatives

??

[D H]

transgalactic,

 

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Please show some work.

[transgalactic]

how to use this hint

{x:f(x)>g(x)} is open

[transgalactic]

this is the only equations i can construct

 

[math]

\mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{f(x_{} ) - f(x_0 )}}{{x - x_0 }}}\limits_{} \\

[/math]

[math]

\mathop {\mathop {\lim }\limits_{x \to x_0 - } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }} = \mathop {\lim }\limits_{x \to x_0 + } \frac{{g(x_{} ) - g(x_0 )}}{{x - x_0 }}}\limits_{} \\

[/math]

 

"if u(x) = f(x)) then there exists an [math]\epsilon > 0[/math] such that u(y) = f(y) for all [math]y \in [x, x + \epsilon)[/math]. So then [math]u'(x)_+ = f'(x)[/math]."

 

i dont know how to apply this this to my question

 

i showed the differentiation equations

i dont know how to continue.

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Merged post follows:

Consecutive posts merged

so u(x) and v(x) around x_0 have to pick different functions

because if one is bigger then the other is smaller.

 

so if f(x)=u(x) then g(x)=v(x) and once again

because f(x) and g(x) are differentiable then we have one sided derivative.

 

i dont know how to write it thematically but here is a try

 

[math]

\forall x_0 \in R

[/math]

[math]

{\rm{ }}\exists \delta {\rm{ > 0 |x - x}}_0 | < \delta {\rm{ }}\mathop {\lim }\limits_{x \to x_0 } u(x) = \mathop {\lim }\limits_{x \to x_0 } v(x)

[/math]

Edited January 31, 2009 by transgalactic
Consecutive posts merged.