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^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
To learn and understanding calculus
To make the function f(x) continuous at x = -3, we need to ensure that the two expressions for f(x) match at x = -3.
First, let's find the value of f(x) when x < -3. According to the given formula,
f(x) = 6x^3 + 17x^2 + 4x + 21 / (x + 3)
For simplicity, let's call this equation 1.
Next, let's find the value of f(x) when x > -3. According to the given formula,
f(x) = 2x^2 - 4x + a
For simplicity, let's call this equation 2.
Since we want to make f(x) continuous at x = -3, we need to find the value of 'a' that makes equation 1 equal to equation 2 at x = -3.
So, we substitute x = -3 into equation 1:
f(-3) = 6(-3)^3 + 17(-3)^2 + 4(-3) + 21 / (-3 + 3)
Simplifying,
f(-3) = -162 + 153 - 12 + 21 / 0
Since we have a division by zero, we have an undefined result. To make the function continuous at x = -3, we cannot have a division by zero.
Therefore, it is not possible to find a single value for 'a' that would make f(x) continuous at x = -3.