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?
mmmh, took me a while to understand what the question is asking.
First of all you will need brackets:
f(x) = (6x^3+17x^2+4x+21)/(x+3)
= (6x^2 - x + 7) , where x≠ -3
= (x+1)(6x-7)
we want this to be the same as
2x^2 - 4x + a
If 6x^2 - x + 7 = 2x^2 - 4x + a
4x^2 + 3x + 7 = a
To make the function continuous at x = -3, we need to ensure that the values of f(x) from both formulas match at x = -3.
First, let's find the value of f(x) using the first formula when x = -3 by substituting x = -3 into the formula:
f(-3) = (6(-3)^3 + 17(-3)^2 + 4(-3) + 21)/(-3 + 3)
= (-162 + 153 - 12 + 21)/0
= 0/0 (undefined)
As you can see, substituting x = -3 into the first formula results in an undefined value. This means that the function is not continuous at x = -3 using the first formula alone.
Now, let's use the second formula to calculate f(x) at x = -3:
f(-3) = 2(-3)^2 - 4(-3) + a
= 18 + 12 + a
= 30 + a
In order for the function to be continuous at x = -3, the values from both formulas should be equal. Therefore:
0/0 = 30 + a
Since dividing by zero is undefined, the only way for the function to be continuous is if the denominator is canceled out. Therefore, we need to find the value of a that makes the denominator (x + 3) cancel out in the first formula.
To achieve this, we need to factorize the numerator of the first formula:
f(x) = 6x^3 + 17x^2 + 4x + 21
Using synthetic division or long division, we find that the numerator can be written as:
(6x + 7)(x^2 + 10x + 3)
So now our original function can be rewritten as:
f(x) = (6x + 7)(x^2 + 10x + 3)/(x + 3)
For the function to be continuous at x = -3, we need to cancel out the (x + 3) term. This can be done by canceling out the (x + 3) term in the denominator with either (6x + 7), (x^2 + 10x + 3), or both.
Since we want to find the value of a, we can cancel out (x + 3) by factoring the (x^2 + 10x + 3) term. Using the quadratic formula, we find the roots of the equation x^2 + 10x + 3 = 0:
x = (-10 ± √(10^2 - 4 * 1 * 3))/(2 * 1)
= (-10 ± √(100 - 12))/2
= (-10 ± √88)/2
= (-10 ± 2√22)/2
= -5 ± √22
So, the two roots are -5 + √22 and -5 - √22. We can factorize the equation as:
x^2 + 10x + 3 = (x - (-5 + √22))(x - (-5 - √22))
Now, since we want to cancel out the (x + 3) term, we need to choose one of the two factors that has x = -3 as a solution. Therefore, we choose:
x - (-5 - √22) = x + 5 + √22
Setting this factor to zero gives us:
x + 5 + √22 = 0
x = -5 - √22
Since x = -3 is not a solution for this factor, we set x = -5 - √22 as the factor that cancels out the (x + 3) term. Therefore, to make the function continuous at x = -3, the value of a should be equal to (-5 - √22).