Math Calculus
posted by URGENT on .
The function f is given by the formula f(x)=6x^3+17x^2+4x+21/x+3
when x<–3 and by the formula
f(x)=2x^2–4x+a
What value must be chosen for a in order to make this function continuous at 3?

6x^3+17x^2+4x+21 = (x+3)(6x^2x+7)
so f(x) > 64 as x > 3
so, we want 2x^24x+a = 64 when x = 3
64 = 18+12+a
a = 34