How to Prove the Sign-Preserving Property for a Continuous Function?

[goodheavens]

how do you prove the sign-preserving property?

it says here that.

If f is continuous at a, and f(a) < 0, then there is an open interval I containing a such that f(x) < 0 for every x in I.

For a proof, simply take the open interval (2f(a),0) for the challenge interval "J" in the definition of continuity.

i don't get it :(

:sorry my previous post was a mistake :)

 

[╔(σ_σ)╝]

Please show us your work.

The idea of the proof is to use the fact that if f is continuous at c and f(c)>0 then there exist a neighbourhood of c such that f(x) >0 .

That neighbourhood is what you have to find.

Try using the definition of continuity.

 

[goodheavens]

i saw a similar problem to this one and I am not sure if this works

using the [tex]\epsilon[/tex] - [tex]\delta[/tex] definition we have

l x-c l < [tex]\delta[/tex] [tex]\Rightarrow[/tex] l f(x) - f(c) l < [tex]\epsilon[/tex]

then they used [tex]\epsilon[/tex] = f(c)/2

lf(x)-f(c)l < f(c)/2

-f(c)/2<f(x)-f(c)< f(c)/2

f(c)/2 < f(x) < 3f(c)/2

f(x) > f(c) > 0 .

but i don't know how [tex]\epsilon[/tex] = f(c)/2

 

[HallsofIvy]

I have restored your second post because it is exactly what you want.

The whole idea of "continuity" is that "if x is close to a, then f(x) is close to f(a)." In particular, if f(c) is positive, and f(x) is close to f(c), then f(x) must be positive also.

Why f(c)/2? If f(c)> 0 then f(c)/2> 0 also so every number between f(c) and f(c)/2 is positive. Find [itex]\delta> 0[/itex] so that if [itex]|x- c|< \delta[/itex] (that is, so that [itex]c- \delta< x< c+ \delta[/itex], then [itex]|f(x)- f(c)|< f(c)/2[/itex]. That is the same as saying that [itex]-f(c)/2< f(x)- f(c)< f(c)/2[/itex] and, adding f(c) to each part, [itex]f(c)/2< f(x)< 3f(c)/2[/itex]. Since f(c)/2 is positive, so is f(x). In other words, f(x) is positive for all x in the interval [itex]c- \delta[/itex] to [itex]c+ \delta[/itex].

 

[goodheavens]

thank you sir

 

[goodheavens]

i forgot to mention that the interval was (a,b)

 

[╔(σ_σ)╝]

Your interval involves c and delta. (a, b) is simply the delta neighbourhood of c.

 

[HallsofIvy]

You cannot determine the interval in advance. Once you have found a "[itex]\delta[/itex]" for whatever [itex]\epsilon[/itex] you choose (as long as [itex]f(c)- \epsilon> 0[/itex]), [itex]a= c- \delta[/itex] and [itex]b= c+ \delta[/itex]

 

[goodheavens]

the interval (a,b) was actually one of the given data.
thank you

 

[╔(σ_σ)╝]

the interval (a,b) was actually one of the given data.
thank you

Well you changed the question!
In the new question b does not exist.

The new question is a minor variation of the previous question. The only different is that c is now a and f(c) <0. This is okay since c was assumped to be abitrary.

The same prove you presented will work but your [itex] \epsilon = \frac{|f(a)|}{2} [/itex].

 

[goodheavens]

Well you changed the question!
In the new question b does not exist.

The new question is a minor variation of the previous question. The only different is that c is now a and f(c) <0. This is okay since c was assumped to be abitrary.

The same prove you presented will work but your [itex] \epsilon = \frac{|f(a)|}{2} [/itex].

yeah cause its the same as in this https://www.physicsforums.com/showthread.php?t=460116 that would be double posting :)

so this is the final proof that i made for the previous question.

Let [tex]\epsilon[/tex] > 0
l x-c l < [tex]\delta[/tex] [tex]\Rightarrow[/tex] l f(x) - f(c) l < [tex]\epsilon[/tex]

Since f(c) > 0. then [tex]\frac{f(c)}{2}[/tex] > 0 . use [tex]\epsilon[/tex] = [tex]\frac{f(c)}{2}[/tex]

l x-c l < [tex]\delta[/tex] [tex]\Rightarrow[/tex] l f(x) - f(c) l < [tex]\frac{f(c)}{2}[/tex]

c - [tex]\delta[/tex] < x < c + [tex]\delta[/tex] [tex]\Rightarrow[/tex] [tex]\frac{-f(c)}{2}[/tex] < f(x) - f(c) < [tex]\frac{f(c)}{2}[/tex]

[tex]\Rightarrow[/tex] [tex]\frac{f(c)}{2}[/tex] < f(x) < [tex]\frac{3f(c)}{2}[/tex]

[tex]\Rightarrow[/tex] f(x) > [tex]\frac{f(c)}{2}[/tex] > 0

Thus, f(c) > 0

if i made a mistake please tell me . thank you

 

[╔(σ_σ)╝]

This is correct.

 

[Char. Limit]

yeah cause its the same as in this https://www.physicsforums.com/showthread.php?t=460116 that would be double posting :)

so this is the final proof that i made for the previous question.

Let [tex]\epsilon[/tex] > 0
l x-c l < [tex]\delta[/tex] [tex]\Rightarrow[/tex] l f(x) - f(c) l < [tex]\epsilon[/tex]

Since f(c) > 0. then [tex]\frac{f(c)}{2}[/tex] > 0 . use [tex]\epsilon[/tex] = [tex]\frac{f(c)}{2}[/tex]

l x-c l < [tex]\delta[/tex] [tex]\Rightarrow[/tex] l f(x) - f(c) l < [tex]\frac{f(c)}{2}[/tex]

c - [tex]\delta[/tex] < x < c + [tex]\delta[/tex] [tex]\Rightarrow[/tex] [tex]\frac{-f(c)}{2}[/tex] < f(x) - f(c) < [tex]\frac{f(c)}{2}[/tex]

[tex]\Rightarrow[/tex] [tex]\frac{f(c)}{2}[/tex] < f(x) < [tex]\frac{3f(c)}{2}[/tex]

[tex]\Rightarrow[/tex] f(x) > [tex]\frac{f(c)}{2}[/tex] > 0

Thus, f(c) > 0

if i made a mistake please tell me . thank you

Weren't you trying to prove that f(x) is greater than 0, using the fact that f(c) is greater than 0? It seems that your final statement should be "Thus, f(x)>0".

 

[goodheavens]

Weren't you trying to prove that f(x) is greater than 0, using the fact that f(c) is greater than 0? It seems that your final statement should be "Thus, f(x)>0".

oh yeah. haha thank you :