Are Square Root Functions Continuous
Continuity of sqrt(x) at x = 0
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Homework Statement
The question is to find two functions (f(x) and g(ten) permit's say) such that they're both NOT continuous at point a but at the same fourth dimension, f(x)+thousand(x) and f(x)yard(x) are continuous.
Homework Equations
The Attempt at a Solution
I was thinking of letting f(x) = ten + [tex]\sqrt{ten}[/tex] and g(10) = x - [tex]\sqrt{x}[/tex], claiming that f(10) and g(10) are non continuous at a = 0. This yields f(x) + g(x) = 2x and f(ten)g(x) = [tex]x^{ii} - ten[/tex]. However, that is the trouble at mitt. Is [tex]\sqrt{x}[/tex] continuous at x = 0? Using the definition of continuity, the limit does NOT be every bit you lot can only notice the limit on i-side (not because the complex plane). However, co-ordinate to my textbook (Stewart), information technology says that all root functions are continuous for every number in its domain. If the latter is the instance, what two functions would satisfy the above? Thank you so much for your aid guys!
Answers and Replies
Roots are continuous. Sqrt(x) is divers for x > 0, so the left limit is not applicable at x = 0.
here'due south a hint : mod functions.
Call back of the functions graphically and what discontinuous functions look like. Depict a bunch of unlike kinds and think of how you maybe be able to add them together and piece them together to make them continuous after adding them.
Well my original tactic was to let
f(ten) = x + (some discontinuous office)
g(x) = x - (some discontinuous function)
so that f(x) + g(10) = 2x and f(x)g(10) = [tex]x^{2}[/tex] - (some discontinuous role)[tex]^{2}[/tex] hoping that the latter would become continuous in one case squared (which is why I wondered if [tex]\sqrt{x}[/tex] was discontinuous at 0 or not). But since that isn't the case, I gauge I've got to find some other mode.
f(ten) = x + (some discontinuous office)
g(x) = x - (some discontinuous function)
so that f(x) + g(10) = 2x and f(x)g(10) = [tex]x^{2}[/tex] - (some discontinuous role)[tex]^{2}[/tex] hoping that the latter would become continuous in one case squared (which is why I wondered if [tex]\sqrt{x}[/tex] was discontinuous at 0 or not). But since that isn't the case, I gauge I've got to find some other mode.
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What kind of discontinuities can you add so that it'd produce a continuous function? The only way I run into is to get rid of them by cancelling them (hence my previous mail). I thought about floor and ceiling functions as someone suggested but what can you lot add together to them to make information technology continuous 
.
.How about step functions?
Hey, here is an instance :
f(x) = 10 + |ten|
yard(x) = x - |ten|
f(x) = 10 + |ten|
yard(x) = x - |ten|
Aren't those continuous in the offset place? f(x) is continuous for all 10 in its domain and is correct-continuous at ten = 0 and one thousand(10) is also continuous with it beingness left-continuous at ten = 0. Otherwise, wouldn't my case with f(x) = x - sqrt(10) and grand(ten) = x + sqrt(x) have worked?
The sqrt thing sort of works, just for the wrong reason. Think of f(10)=ane if x>=0 and f(x)=0 if x<0. Let thou(x)=1-f(x).
Suggested for: Continuity of sqrt(x) at x = 0
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Are Square Root Functions Continuous,
Source: https://www.physicsforums.com/threads/continuity-of-sqrt-x-at-x-0.192077/
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