The function f is given by the formula f(x)=6x^3+17x^2+4x+21/x+3

Question

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?

Answer 1

6x^3+17x^2+4x+21 = (x+3)(6x^2-x+7)
so f(x) -> 64 as x -> -3

so, we want 2x^2-4x+a = 64 when x = -3
64 = 18+12+a
a = 34

Answer 2

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