Show that max(f,g) is continious?is my answer wrong?

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  • #1
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Homework Statement


let f and g are continuous functions show that max(f,g) is continious


The Attempt at a Solution


the answer i gave is this
maxf(g)=h(x)
h(x)=f(x)==>f(x)-g(x)>0
h(x)=g(x)==>f(x)-g(x)≤0
thus h is continious

now in the text book they defined max with the normal definition
max=lf(x)-g(x)l+f(x)+g(x)/2
is my answer wrong?
 
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Answers and Replies

  • #2
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Homework Statement


let f and g are continious functions show that max(f,g) is continious


The Attempt at a Solution


the answer i gave is this
maxf(g)=h(x)
h(x)=f(x)==>f(x)-g(x)>0
h(x)=g(x)==>f(x)-g(x)≤0
thus h is continious
Why does this imply that h is continuous?
 
  • #3
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I thought it's obvious
if f(x)>g(x) then h(x) = f(x) f continious then h continious
if f(x)<=g(x) then h(x)=g(x) g gotinious thus h is continious
Am i doing something wrong here?(spivak's calculus questions are quiet hard )
Edit : i wasn't supposed to study cases?
 
  • #4
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Homework Statement


let f and g are continious functions show that max(f,g) is continuous
"Continious" is not a word - the one you mean is continuous.

The Attempt at a Solution


the answer i gave is this
maxf(g)=h(x)
Presumably you mean max(f, g) = h(x)
h(x)=f(x)==>f(x)-g(x)>0
h(x)=g(x)==>f(x)-g(x)≤0
The above assume that f > g for all x in the domain or that g ≤ f for all x in the domain. What happens if one function is greater than the other in some places, but not in others?
thus h is continious

now in the text book they defined max with the normal definition
max=lf(x)-g(x)l+f(x)+g(x)/2
is my answer wrong?
For absolute value, use |, not l (lower-case "ell"). Also, what gets divided by 2? As you wrote it, only g(x) is divided by 2. If that's not what you meant, use parentheses.
 
  • #5
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3,283
This looks like the following proof:

Consider ##f(x) = 1## for all ##x##, ##g(x) = -1## for all ##x## and take

[tex]h(x) = \left\{\begin{array}{l} 1~\text{if}~x<0\\ -1~\text{if}~x\geq 0\end{array}\right.[/tex]

If ##h(x)<0##, then ##h(x) = g(x)## and ##g## is continuous.
If ##h(x)>0##, then ##h(x) = f(x)## and ##f## is continuous.
Thus ##h## is continuous.

This is essentially what you did. But the conclusion that ##h## is continuous is false of course.
 
  • #6
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The above assume that f > g for all x in the domain or that g ≤ f for all x in the domain. What happens if one function is greater than the other in some places, but not in others?
i was missing this part thank you
micromass made me laugh a bit at my "proof" :).
thanks guys .
 

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