# Continuity Proof

#### seams192

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.

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#### Drexel28

MHF Hall of Honor
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:

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

#### seams192

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

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle \max\{f,g\}=\frac{f+g+|f-g|}{2}$$, can you show that's continuous by using some of your known theorems?

#### 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?

#### Drexel28

MHF Hall of Honor
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 (Worried)

#### Opalg

MHF Hall of Honor
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.

• HallsofIvy

#### rmorin

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 $$\displaystyle x$$ (element of domain) is point internal, and demostrate than is continiuos

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

#### mabruka

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

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

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

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

• seams192

#### Pinkk

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