Proving f(x) = 9x - 7 is continuous at x = 2

[Jeff Ford]

I am working to prove that this function is continuous at $x = 2$

$${f(x) = 9x - 7}$$

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

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

This is as far as I got

$$\vert {f(x) – f(a)} \vert < \epsilon$$
$$\vert {9x – 7} \vert < \epsilon$$

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

A push in the right direction would be appreciated.

 

[quasar987]

Wow. First of all, I suggest you use the more conventionnal sign "-" for minus, instead of 8211; .

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.

 

[Jeff Ford]

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.