If f is continuous at c and f(c)>1

  • Thread starter H2Pendragon
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In summary: But to prove the statement, you need to show that there exists an r > 0 such that for all x in the ball around c with radius r, f(x) is greater than 1. This can be done by choosing a small enough value of epsilon, such that when you add it to f(c), you get a value greater than 1. This shows that there is a positive distance between f(c) and 1, meaning that there exists an r that satisfies the statement. Therefore, in summary, by choosing a small enough value of epsilon, we can show that there exists an r > 0 such that for all x in the ball around c with radius r, f(x) is greater than 1. This proves
  • #1
H2Pendragon
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Homework Statement



Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

Let f be continuous at x=c and f(c) > 1

Show that there exists an r > 0 such that [tex]\forall x \in B(c,r) \bigcap D : f(x) > 1[/tex]

Homework Equations



[tex]\forall \epsilon > 0, \exists r > 0, \forall x \in B(c,r) \bigcap D \Rightarrow |f(x) - f(c)| < \epsilon [/tex]

f(c) > 1

The Attempt at a Solution



I'd put something here but I don't really have any attempts. I originally thought of using the limit rule that lim f(x) = f(c) but that doesn't work since c could be an isolated point.

If I used a graph it just seems so obvious, but I have to use words to prove it. I tried also to use a contradiction, but using just what I put down in (2) didn't lead to any direct contradiction that I could see.

A nudge in the right direction would be helpful.
 
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  • #2
If in doubt try an example.

Let's say for the sake of argument that f(c)=2. What choice of epsilon, and r, from 2. above will mean that f(x) > 1 in the ball around c?
 
  • #3
-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.
 
  • #4
MathematicalPhysicist said:
-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.

by this reasoning couldn't I just claim that f(c) = 1 + e?
 
  • #5
Yes, you can do that.
 

FAQ: If f is continuous at c and f(c)>1

1. What does it mean for a function to be continuous at a certain point?

Being continuous at a point means that the function is defined and has a value at that point, and as the input values get closer to that point, the output values also get closer to the same value. In other words, there are no sudden jumps or breaks in the function at that point.

2. How can you determine if a function is continuous at a given point?

A function is continuous at a point if the limit of the function as the input approaches that point is equal to the output value at that point. This can be checked by evaluating the limit mathematically or graphically.

3. What does it mean for f(c) to be greater than 1 in the statement "If f is continuous at c and f(c)>1"?

In this statement, f(c) refers to the output value of the function at the specific point c. If this output value is greater than 1, it indicates that the function has a value greater than 1 at that point.

4. Can a function be continuous at a point but not have a value greater than 1 at that point?

Yes, a function can be continuous at a point even if the output value at that point is not greater than 1. Continuity at a point only requires that the limit of the function approaches the output value at that point, not that they are equal or greater than each other.

5. How does the statement "If f is continuous at c and f(c)>1" relate to real-world scenarios?

This statement can be applied to real-world scenarios, such as a car's speed on a journey. If f represents the speed function, then f(c) would be the speed at a specific time c. If the car is continuously moving at a speed greater than 1 (e.g. 60mph), then the function is continuous at that point. However, the function could also be continuous if the car's speed is less than 1 (e.g. 30mph) as long as it does not have any sudden changes or stops at that point.

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