What value must be chosen for a in order to make this function continuous at -3?

Question

The function f  is given by the formula: 

 

f(x)= (6x^3+18x^2+9x+27)/(x+3) when x < -3

 

and by the formula: f(x) = 5x^2+6x+a when -3 ≤ x

 

What value must be chosen for a in order to make this function continuous at -3?

 

a= ?

Answer 1

First see if you can factor the numerator of the first expression for f(x).  If you can, and if one of the factors is x+3, then the other factors will constitute a function that is identical to f(x) when x < -3.

 

6x³+18x²+9x+27 = 3(2x³ + 6x² + 3x + 9) = 3(2x² + 3)(x + 3).  (NOTE:  I did this factoring by doing long division of 2x³ + 6x² + 3x + 9 by x + 3.)

 

So when x < -3, the function g(x) = 3(2x² + 3) is identical to f(x).

 

So the limits of the two functions as x --> -3 from below are the same.  So in order for f(x) to be continuous at x = -3, a necessary condition is that f(-3) has to be equal to g(-3).  g(-3) = 3(18 + 3) = 63.

 

So you want f(x) to be continuous at -3, then 5x²+6x+a has to be 63 when x = -3.  That means that:

 

5(-3)²+6(-3)+a = 63

45 - 18 + a = 63

27 + a = 63

a = 36