Proving a function is continuous

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SUMMARY

This discussion focuses on proving the continuity of the function f(x) = 9x - 7 at x = 2. The key requirement is to demonstrate that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |f(x) - f(2)| < ε. The limit as x approaches 2 is calculated as f(2) = 11, leading to the expression |9x - 18| = 9|x - 2| < ε, which establishes the continuity of the function at that point.

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  • Familiarity with the ε-δ definition of continuity
  • Basic algebraic manipulation skills
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Jeff Ford
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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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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.
 
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.
 

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