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.

 

[BerkMath]

You are making this way to difficult. There are two equivalent definitions of continuity. The epsilon-delta one which you are attempting to use, which will work when done correctly, and the limit definition. The latter works well when proving continuity of functions such as yours 9x-7. Stated briefly, a function f is continuous at x=a if for every sequence xn converging to a lim f(xn)=f(a) (n->inf). From here it is quickly seen that your function is continuous at x=2. Let xn be a sequence that converges to 2, xn->2 as n->inf, then lim f(xn)=f(2) (n->inf), this is just from definition of limits, however, f(a) is simply f(2). Therefore they are equivalent, implying f(x)=9x-7 is continuous at x=2.