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- anonymous

help here

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- anonymous

@Michele_Laino

- anonymous

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- anonymous

please help with question number 5b

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## More answers

- 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

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

in other words those three statements above allow us to say that max(x,y) and min(x,y) are continuous functions

- ganeshie8

can w use epsilon-delta definition to prove the statements ?

- Michele_Laino

I don't know, since f and g are generic functions

- Michele_Laino

yes! we can prove those last three statements using epsilon-delta

- 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

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

ok!

- 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

ok!

- 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

greater*

- anonymous

is the proving still going on?

- 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

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

Please double check.. it might contain errors as im really bad with real analysis...

- Michele_Laino

that's right! @ganeshie8

- Michele_Laino

I was thinking to your method :)

- ganeshie8

xD

- Michele_Laino

now you have two different solutions @GIL.ojei

- anonymous

thank you sirs but is that all from you @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

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

@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

@ganeshie8

- anonymous

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