Proving a function is continuous

In summary, to prove that the function f(x) = 9x - 7 is continuous at x = 2, it is necessary to show that |f(x) - f(a)| < e and |x - a| < d. However, finding a relationship between |x - 2| and e to determine an appropriate d has proven difficult. A suggestion to use the limit as x approaches a and the given values of a and f(a) is provided.
  • #1
Jeff Ford
155
2
I am working to prove that this function is continuous at [itex] x = 2 [/itex]

[tex] {f(x) = 9x - 7} [/tex]

To do this I know that I have to show that [itex] \vert f(x) – f(a) \vert < \epsilon [/itex] and that [itex] \vert x - a < \delta \vert [/itex]

I tried to come up with a relationship between [itex] \vert x - 2 \vert [/itex] and [itex] \epsilon [/itex] so I could get an appropriate number to choose for [itex] \delta [/itex]

This is as far as I got

[tex] \vert {f(x) – f(a)} \vert < \epsilon [/tex]
[tex] \vert {9x – 7} \vert < \epsilon [/tex]

I’m stuck. All of the examples the text shows give equations where it is easy to factor out the [itex] \vert {x - a} \vert [/itex] term.

A push in the right direction would be appreciated.
 
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  • #2
Wow. First of all, I suggest you use the more conventionnal sign "-" for minus, instead of 8211; :wink:.

To show a function is continuous at 'a', you must how that the limit as x approaches a is f(a). Here, a = 2 and f(2) = 9(2)-7 = 11. So given e>0, we must find 'd' such that 0<|x-2|<d ==> |9x-7 - 11|=|9x-18|=9|x-2|<e.

Mmmh.
 
  • #3
Thanks. Latex was being weird yesterday. I'm not sure why it put the 8211; in there in place of some - but not all. Weird.
 

What does it mean for a function to be continuous?

A function is continuous if its graph can be drawn without lifting the pencil from the paper. In other words, there are no gaps or breaks in the graph and the function has a smooth, continuous shape.

What are the three requirements for a function to be continuous?

A function is continuous if it satisfies the following three requirements:
1. The function is defined at the point in question.
2. The limit of the function at that point exists.
3. The value of the function at that point is equal to the limit.

How do you prove a function is continuous at a specific point?

To prove a function is continuous at a specific point, you must show that it satisfies the three requirements for continuity. This can be done by evaluating the function at the point and taking the limit as the input approaches the point in question. If the limit exists and is equal to the function's value at that point, the function is continuous at that point.

What is the difference between continuity and differentiability?

Continuity and differentiability are related but distinct concepts. A function is continuous if its graph has no breaks or gaps, while a function is differentiable if it has a well-defined derivative at every point. In other words, continuity refers to the overall shape of a function, while differentiability refers to the steepness of the function's graph.

Can a function be continuous but not differentiable?

Yes, it is possible for a function to be continuous but not differentiable. This can occur at points where the function has a sharp turn or corner, as the derivative does not exist at these points. However, a function cannot be differentiable if it is not continuous, as the three requirements for continuity must be satisfied in order for a function to be differentiable.

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