help hard calculus question

Question

a. graph the function f(x) = { x^3+x^2-12x/x-3, if x does not equal 3

{ 15, if x=3

b. determine f(3)

c. determine lim x-> 3^- f(x), lim x-> 3^+ f(x) , and lim x-> 3 f(x)

d. is f(x) continuous? explain

Answer 1

f(x) = (x³ + x² - 12x) / (x - 3) = x(x² + x - 12) / (x - 3) = x(x + 4)(x - 3) / (x - 3)

So f(x) = x(x + 4) for all x such that x ≠ 3. (We cancel the (x-3) from the numerator and denominator for x-3≠0.)

So f(x) will have a removable discontinuity (hole) at x = 3, rather than a vertical asymptote, which happens if the denominator is NOT a factor of the numerator. To determine the y-coordinate of the hole, and thus the limits of the function as x -> 3, we simply plug x = 3 into what remains after cancelling:

g(x) = x(x + 4)

g(3) = 3(3 + 4) = 21

So f(x) graphs like the concave up parabola y = x(x + 4) except it has a hole at (3,21) and a point defined at (3,15). This makes the function discontinuous there, since the lim_x->3 f(x) = 21 but f(3) = 15, thus the limit does not = the value of the function at x = 3.

Btw, lim_x→3- f(x) = lim_x→3+ f(x) = lim_x→3 f(x) = 21