Continuity Question in calculus BA for explanation, not answer

Define g(9) for the given function so that it is continuoous at x=9:

g(x) = (6^2 - 486) / (6x-54)

I need to know how to approach these types of problems, If someone could walk me through it, your help would be greatly appreciated. Will be quick to award BA! Thanks in advance

You must have meant g(x) = (6x^2 - 486) / (6x - 54), since what you've written can't be made continuous at x=9.

There are two approaches to this type of problem. The first uses factoring. Your question can be written as g(x) = (x^2 - 81)/(x-9). In this form, it's obvious that the numerator factors as (x-9)(x+9), so
g(x) = (x-9)(x+9)/(x-9). Now you can cancel the x-9 in the numerator and denominator to get g(x) = x+9. This is valid *whenever x is not 9*. When x is 9, the original function has a divide by 0 error. Since x+9 is continuous and agrees with g(x) everywhere except at 9, you have to define g(9) = 9+9 = 18.

The second approach is to use L'Hopital's rule. That evaluates the limit as x approaches 9, and you define g at 9 so that it agrees with this limit. L'Hopital's rule only applies since x^2 - 81 and x-9 are both 0 at x=9, so you get the indeterminate form 0/0. The rule here gives
lim x->9 g(x)
= lim x->9 (x^2 - 81)/(x-9)
= lim x->9 (2x) / 1
= 2*9 = 18.

Are you sure that it is g(x) = (6^2 - 486) / (6x-54) and not g(x) = (6x^2 - 486) / (6x-54) ?