anonymous
  • anonymous
help here
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
anonymous
  • anonymous
please help with question number 5b

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Michele_Laino
  • Michele_Laino
here we have to keep in mind that: 1) the addition is a continuous function on R, namely R x R ---> R 2) the opposite value function is continuous in R, namely R--->R 3) the absolute value function is continuous in R, namely R---> R+ of course all those properties can be proved
Michele_Laino
  • Michele_Laino
so, using the theorem which states that the composition of two continuous functions is again a continuous function and the fact that the previous functions, namely max(x,y) and min(x,y) are continuous functions, then we get your thesis
Michele_Laino
  • Michele_Laino
in other words those three statements above allow us to say that max(x,y) and min(x,y) are continuous functions
ganeshie8
  • ganeshie8
can w use epsilon-delta definition to prove the statements ?
Michele_Laino
  • Michele_Laino
I don't know, since f and g are generic functions
Michele_Laino
  • Michele_Laino
yes! we can prove those last three statements using epsilon-delta
Michele_Laino
  • Michele_Laino
nevertheless , since we can view this function: f(x)+g(x) as the composition of the functions: x---> (f(x),g(x)) and the addition function, then we can apply the theorem of composition of continuous function, in order to say that the function: f(x)+g(x) is a continuous function when also f and are continuous
ganeshie8
  • ganeshie8
correct me if im wrong michele Since \(f\) and \(g\) are continuous at \(a\), we have : \[\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < |x -a| < \delta_f \implies |f(x) - f(a)| < \varepsilon\] \[\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < |x -a| < \delta_g \implies |g(x) - g(a)| < \varepsilon\] let \(\delta_h=\min{\{\delta_f,~\delta_g\}}\) and \(\delta_k=\max{\{\delta_f,~\delta_g\}}\)
Michele_Laino
  • Michele_Laino
ok!
ganeshie8
  • ganeshie8
we want to show below : \(|x-a|\lt \delta_h \implies |h(x)-h(a)| \lt \epsilon\) \(|x-a|\lt \delta_k \implies |k(x)-h(a)| \lt \epsilon\)
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
yes! I think that we can prove the statement using the epsilon-delta, if we consider these 2 cases, namely when f greter or equal to g, and when f less or equal to g. In those cases, we have: h(x)= f(x) and k(x) = g(x), and vice versa for second case, and by hypothesis both f and g are continuous
Michele_Laino
  • Michele_Laino
greater*
anonymous
  • anonymous
is the proving still going on?
ganeshie8
  • ganeshie8
I think the same \(\delta\) works for both \(h(x)\) and \(k(x)\). Here is the complete proof : Since \(f\) and \(g\) are continuous at \(a\), we have : \[\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < |x -a| < \delta_f \implies |f(x) - f(a)| < \varepsilon ~~(\color{red}{\star}) \] \[\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < |x -a| < \delta_g \implies |g(x) - g(a)| < \varepsilon ~~(\color{red}{\diamond}) \] Let \(\delta_h=\delta_k=\min{\{\delta_f,~\delta_g\}}\) and suppose that \(f(a)\ge g(a)\), then : \(|x-a|\lt \delta_h \implies\) \(\begin{align}|h(x)-h(a)| &=|\max\{f(x),g(x)\}-\max\{f(a),g(a)\}|\\~\\ &=|\max\{f(x),g(x)\}-f(a)\}|\end{align}\) Case1 : \(f(x)\ge g(x)\) \(|\max\{f(x),g(x)\}-f(a)\}|=|f(x)-f(a)|\lt \epsilon\) by \((\color{red}{\star})\) Case2 : \(g(x)\ge f(x)\) \(|\max\{f(x),g(x)\}-f(a)\}|=|g(x)-f(a)| \lt |g(x)-g(a)|\lt \epsilon\) by \((\color{red}{\diamond})\) similarly we can prove the other statement by considering two cases
Michele_Laino
  • Michele_Laino
referring to my proof we have finished. So reassuming, in order to prove your statement, we need of the subsequent properties: 1) the addition is a continuous function on R, namely R x R ---> R 2) the opposite value function is continuous in R, namely R--->R 3) the absolute value function is continuous in R, namely R---> R+ furthermore, using the theorem of composition of continuous functions, and the fact that this function: x--->(f(x),g(x)) is continuous, then we get your thesis
ganeshie8
  • ganeshie8
Please double check.. it might contain errors as im really bad with real analysis...
Michele_Laino
  • Michele_Laino
that's right! @ganeshie8
Michele_Laino
  • Michele_Laino
I was thinking to your method :)
ganeshie8
  • ganeshie8
xD
Michele_Laino
  • Michele_Laino
now you have two different solutions @GIL.ojei
anonymous
  • anonymous
thank you sirs but is that all from you @ganeshie8
ganeshie8
  • ganeshie8
I think what I had there proves that \(h(x)\) is continuous at \(a\). I'm hoping you can mimic the same easily for proving continuity of \(k(x)\) too...
anonymous
  • anonymous
I WILL BUT IT WOULD BE MORE EASIER FOR ME IF YOU COMPLETE IT IN OTHER NOT TO HAVE ERRORS, COS DON'T KNOW HOW TO USE THE SYMBOLS IN THIS SITE
anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
(... continued) \(|x-a|\lt \delta_k \implies\) \(\begin{align}|k(x)-k(a)| &=|\min\{f(x),g(x)\}-\min\{f(a),g(a)\}|\\~\\ &=|\min\{f(x),g(x)\}-g(a)\}|\end{align}\) Case1 : \(f(x)\ge g(x)\) \(|\min\{f(x),g(x)\}-g(a)\}|=|g(x)-g(a)|\lt \epsilon\) by \((\color{red}{\star})\) Case2 : \(g(x)\ge f(x)\) \(|\min\{f(x),g(x)\}-g(a)\}|=|f(x)-g(a)| \lt |g(x)-g(a)|\lt \epsilon\) by \((\color{red}{\diamond})\) \(\blacksquare\)
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous

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