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Product of two continous functions 
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#1
Feb1008, 07:34 PM

P: 6

There is a proof in my book that asks us to prove that the product of two continuous functions is continuous. If anyone could help please reply back, thanks!



#2
Feb1008, 08:12 PM

P: 104

If h(x) is undefined at some point c (and thus not continuous), will it ever have a product that is defined at that point?



#3
Feb1108, 07:18 AM

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It is not clear to me what SticksandStones response has to do with the question!
im2fastfouru, you know, of course, that the product, fg, will be continuous at a as long as the lim, as x goes to a, of f(x)g(x) is equal to f(a)g(a). Therefore, you must prove that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if xa< [itex]\delta[/itex], then f(x)g(x) f(a)g(a)< [itex]\epsilon[/itex]. Using, of course, the fact that f and g are both continuous at a, that is, that given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if xa< [itex]\delta[/itex] such that f(x) f(a)< [itex]\epsilon[/itex] and similarly for g. I would recommend you look at f(x)g(x) f(x)g(a)+ f(x)g(a) f(a)g(a). After all, your book asked you to prove this, not us! 


#4
Feb1108, 10:43 AM

P: 104

Product of two continous functions
I thought he was asking something entirely different; I apologize.



#5
Feb1108, 06:03 PM

HW Helper
P: 2,263

hint
write f(x+h)=f(x)+[f(x+h)f(x)] g(x+h)=g(x)+[g(x+h)g(x)] note f(x+h)f(x)<eps1 g(x+h)g(x)<eps2 f(x+h)f(x),g(x+h)g(x)<eps=max(eps1,eps2) also recall a+b+c<a+b+c 


#6
Feb1208, 01:11 AM

P: 1,635

For the OP: do you see how to go now? Like Halls said, you should give some effort on your own next time, if you want to recieve more help. People here won't just give us answers(including me since i often ask for help). 


#7
Feb1208, 05:10 AM

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lurflurf may has misread "continuous" as "differentiable".



#8
Feb1208, 08:41 AM

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the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.
to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small. 


#9
Feb1808, 05:51 PM

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P: 2,263

lim_h>0 (f(x+h)f(x))=0 where (for picking nits) we accept as implicit f(x) exist and limit_h>0 f(x+h) exist we desire to show that if f(x+h)f(x)<eps1 g(x+h)g(x)<eps2 then f(x+h)g(x+h)f(x)g(x)<eps write f(x+h)=f(x)+[f(x+h)f(x)] g(x+h)=g(x)+[g(x+h)g(x)] thus f(x+h)g(x+h)f(x)g(x)=[f(x+h)f(x)]g(x)+f(x)[g(x+h)g(x)]+[f(x+h)f(x)][g(x+h)g(x)] <[f(x+h)f(x)g(x)+f(x)g(x+h)g(x)+f(x+h)f(x)g(x+h)g(x)] <eps1g(x)+f(x)eps2+eps1*eps2 letting eps1=(eps/3)/g(x) if g(x)>0 eps2=(eps/3)/max(eps1,f(x)) if f(x)>0 we easily see the result is true 


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