How would you "remove the discontinuity" of f? In other words, how would you define
f(5) in order to make f continuous at 5?

f(x) = [(x^2 -2x -15)/x-5]

Question

How would you "remove the discontinuity" of f? In other words, how would you define 
f(5) in order to make f continuous at 5?

f(x) = [(x^2 -2x -15)/x-5]

anonymous · Answer

plug the hole at x = 5

anonymous · Answer

how would you do that, that is what i am confused on

anonymous · Answer

you first need to find the value of the limit at x=5
so$$\lim_{x \rightarrow 5} {{x^2-2x-15}\over{x-5}}$$
$$\lim_{x \rightarrow 5} {(x-5)(x+3)\over(x-5)}$$
after cancelling, the limit is 8
therefore you need to describe you function as
$$f(x)=|^{x^2-2x-15 , x \neq 5}_{8, x=5} $$

anonymous · Answer

ah i see, i didnt realize i just had to find the limit..i was over thinking, thank you for your help!