2 discontinuity problems

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hsiwin3

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Sep 10, 2009
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I have two problems i need help with please:

(1) Give an example of functions f(x) and g(x) such that f(g(x)) is con-
tinuous everywhere but g(x) has at least one discontinuity.

---can someone help me find an f(x) that would make f(g(x)) continuous is g(X)=1/x?



(2) Let f be the function defined as follows:
f(x) =
(x x is rational
(-x x is irrational

Show that f(x) is continuous at x=0, but discontinuous for all other real numbers x.

---i don't understand this one at all. i don't understand what the graph would look like.


any help would be greatly appreciated !
 
1) Think along the lines of rational functions p(x)/q(x) such that 1/q(x) has a discontinuity (I'd choose a single, removable one) and 1/p(x) has has none. Then let g(x) = p(x)/q(x), f(x) = 1/x. Then observe what happens to f(g(x)).


(2) Seriously? This is a big leap from the last question in terms of material IMO. If |x| < delta then since f(x) is either x or -x, |x| = |f(x)|< delta. So chooose delta=epsilon. To show the other part, let p be a positive real number and choose epsilon as p/2. Assume you can find a delta such that |x-p| < delta means that |f(x)-f(p)|<epsilon. We may assume delta < p (if it is not, make it so).

There are a couple cases: If p is rational, then f(p)=p. Hence, |x-p| = |x-f(p)| < delta. There is an irrational number x' within this interval, so we have |x'-f(p)| < delta => |f(x')-f(p)| = |-x'-f(p)| = |x'+f(p)| < epsilon = p/2.

Now, we know f(p)=p and since we chose delta to insure p-delta > 0, we know x'>0 since it is within delta of p. This implies|x'+f(p)| = x'+f(p). Therefore we have |x'+f(p)| = x'+f(p) > f(p) = p > p/2 = epsilon. and by the above argument, |x'+f(p)| < p/2. See the contradiction?

Now try the case that p is irrational, and see if you can make modifications for p < 0.
 
do you know what a rational function is? look that up first.
 
yes i do. for p(x)/q(x) do you mean something like (x^2+x-20)/(x^2-3x-18) ?
 
yup. but that choice for p will not suffice. you need 1/p(x) to have no discontinuities and if you think about the rest of what i wrote you'll see why.
 
ok so let me see if i have this right..

if i have f(x)=1/sqrt(x) and g(x)=1/(x-1)

that would work?
 
if you consider "everywhere" to mean "everywhere on its domain" then yes. you can stick any polynomial into a square root function and get that result... If you want continuous on R, better stick with rational functions (sqrt(x) is not a polynomial).

if p(x) is a polynomial such that p(x) is not zero for any x, then 1/p(x) is continuous for all x. therefore if q(x) is a polynomial such that 1/q(x) is discontinuous then so is p(x)/q(x). so i let f(x)=p(x)/q(x) and g(x)=1/x. its up to you to find these polynomials and show that f(g(x)) is continuous everywhere, as if i did any more i'll have just done your homework for you.

what i wrote is merely a suggestion, they need not be rational functions
 
hsiwin3 said:
… (1) Give an example of functions f(x) and g(x) such that f(g(x)) is con-
tinuous everywhere but g(x) has at least one discontinuity …

… [could] g(x)=1/x ? … I think so.
Click to expand...


:idea: (Is this cheating?)

f(x) = 1/x

g(x) = 1/x

 
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