Proving f-g is Continuous & f Inverse Existence

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SUMMARY

The discussion focuses on proving that the difference of two continuous functions, f and g, denoted as f-g, is also continuous. It establishes that by applying the triangle inequality, |f(x) - g(x) - (f(a) - g(a))| can be bounded by the sum of the absolute differences |f(x) - f(a)| and |g(x) - g(a)|, both of which can be made less than any ε due to the continuity of f and g. Additionally, it concludes that while f is continuous, the existence of its inverse depends on the function's domain; for example, y=x² is continuous but does not have an inverse over all real numbers.

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Homework Statement


a) If f and g are continuous functions how do you show f-g is continous?

b) if f is continuous then f inverse exists?


The Attempt at a Solution


a) Do you look at the sets that f and g maps points in the domain into?

b) I think so.
 
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(a) How about looking at [itex]|f(x)- g(x)- (f(a)- g(a))|\le |f(x)- f(a)|+ |g(x)-g(a)| as just about every calculus book does? Since f is continuous (at a) you can make that first absolute value less than any [itex]\epsilon[/itex], since g is continuous (at a), you can make the second absolute value less than any epsilon.<br /> <br /> (b) What about y= x<sup>2</sup> with domain all real numbers? Is it continous? Does it have an inverse? What do the DEFINITIONS of "continous" and "inverse function" have to do with each other?[/itex]
 
For y=x^2 it has an inverse for [0,infinity) but not all the real numbers, so no for b)
 

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