1. ## Continuity Proof

Sorry reposted:

Let f(x) and g(x) be fcns de fned on an interval I and let h(x) = max(f(x); g(x)) and j(x) = min(f(x); g(x)). h(x) and
j(x) are continuous on I. Show f(x) and g(x) are continuous on I or give a counterexample.

2. Originally Posted by seams192
Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.

Any ideas on how to structure this proof using contradiction?

Thanks!
How much do you know about analysis?
Hint:

Spoiler:

$\left(f\vee g\right)^{-1}((a,\infty))=\left\{x\in I:f(x)>a\right\}\cup\left\{x\in I:g(x)>a\right\}$

3. I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.

4. Originally Posted by seams192
I know analysis at an introduction course level. I know about delta-epsilon, continuity definitions, IVT, etc.
Hmm. Alright, so I guess the easiest way to say it then would be that $\max\{f,g\}=\frac{f+g+|f-g|}{2}$, can you show that's continuous by using some of your known theorems?

5. Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?

6. Originally Posted by seams192
Well, we know that the function h(x) is continuous, so the max(f,g) would be continuous. I know that there are rules for algebraic combinations of continuous functions being continuous, so I'm wondering if the max function can be worked backwards to prove f and g continuous that way. Is that where you were going with this?
Oh, wow! I'm sorry, I completely misread the question. Please forgive me, I have to go. I hope another member can help you. Once again, I'm sorry

7. Originally Posted by seams192
Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.

Any ideas on how to structure this proof using contradiction?
A better idea would be to look for a counterexample, perhaps by taking f to be a function that only takes the values 0 and 1, with a jump discontinuity at some point. Then think about how to construct g so as to make h and j continuous.

8. Originally Posted by seams192
Let f(x) and g(x) be functions defi ned on an open interval I and let h(x) = max{f(x); g(x)} and j(x) = min{f(x); g(x)}. Suppose that h(x) and
j(x) are continuous on I. Either show that f(x) and g(x) are also continuous on I or else give a counterexample.

Any ideas on how to structure this proof using contradiction?

Thanks!
take two parts, first suposse than $x$ (element of domain) is point internal, and demostrate than is continiuos

second suposse that $x$ satisfy f(x)=g(x) and probe that is continius.

9. Or simply take these two easy functions on [0,1]

f(x)=1 if $0\leq x<\frac{1}{2}$
-1 if $\frac{1}{2} \leq x \leq 1$

g(x)= -1 if $0\leq x <\frac{1}{2}$
1 if $\frac{1}{2} \leq x \leq1$

both are discontinuous but max and min functions h,j are constant hence continuous

10. Uh, don't delete the question just because it has been answered.

11. sorry pinkk...I put it back up