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]

Given f(x) = [(x^2 -2x -15)/x-5], notice it is a rational function

But the numerator x^2 -2x -15 could be factored, which yields

(x – 5)(x + 3)

Therefore f(x) = (x – 5)(x + 3)/ (x-5)

Cancelling x-5, f(x)  = x + 3

In this way, the f(x) is continuous at any point, and is basically a line.

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph