Answer :
To find the values of [tex]\( x \)[/tex] where the function [tex]\( \frac{x^2 + 2x + 3}{x^2 - x - 12} \)[/tex] is discontinuous, we need to determine where the denominator of the fraction is equal to zero. A function with a rational expression is discontinuous where its denominator equals zero because the function value is undefined at these points. Thus, we need to solve:
[tex]\[ x^2 - x - 12 = 0 \][/tex]
We solve this quadratic equation to find the roots. The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -12 \)[/tex].
To solve [tex]\( x^2 - x - 12 = 0 \)[/tex], we can factor it. We need to find two numbers that multiply to [tex]\(-12\)[/tex] (the constant term) and add up to [tex]\(-1\)[/tex] (the coefficient of the linear term [tex]\(x\)[/tex]).
Considering the factors of [tex]\(-12\)[/tex], we have pairs: [tex]\((1, -12)\)[/tex], [tex]\((-1, 12)\)[/tex], [tex]\((2, -6)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((3, -4)\)[/tex], [tex]\((-3, 4)\)[/tex]. Among these pairs, the pair that adds up to [tex]\(-1\)[/tex] is [tex]\((-4, 3)\)[/tex].
So, we can factor the equation [tex]\( x^2 - x - 12 \)[/tex] as:
[tex]\[ (x - 4)(x + 3) = 0 \][/tex]
Setting each factor equal to zero gives the solutions:
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
Thus, the values of [tex]\( x \)[/tex] where [tex]\(\frac{x^2 + 2x + 3}{x^2 - x - 12}\)[/tex] is discontinuous are [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
Among the choices given:
- [tex]\( x = -1 \)[/tex]
- [tex]\( x = -2 \)[/tex]
- [tex]\( x = -3 \)[/tex]
- [tex]\( x = -4 \)[/tex]
The value from this list that is a discontinuity for the function is:
[tex]\[ x = -3 \][/tex]
Therefore, the correct answer is:
[tex]\[ x = -3 \][/tex]
[tex]\[ x^2 - x - 12 = 0 \][/tex]
We solve this quadratic equation to find the roots. The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -12 \)[/tex].
To solve [tex]\( x^2 - x - 12 = 0 \)[/tex], we can factor it. We need to find two numbers that multiply to [tex]\(-12\)[/tex] (the constant term) and add up to [tex]\(-1\)[/tex] (the coefficient of the linear term [tex]\(x\)[/tex]).
Considering the factors of [tex]\(-12\)[/tex], we have pairs: [tex]\((1, -12)\)[/tex], [tex]\((-1, 12)\)[/tex], [tex]\((2, -6)\)[/tex], [tex]\((-2, 6)\)[/tex], [tex]\((3, -4)\)[/tex], [tex]\((-3, 4)\)[/tex]. Among these pairs, the pair that adds up to [tex]\(-1\)[/tex] is [tex]\((-4, 3)\)[/tex].
So, we can factor the equation [tex]\( x^2 - x - 12 \)[/tex] as:
[tex]\[ (x - 4)(x + 3) = 0 \][/tex]
Setting each factor equal to zero gives the solutions:
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]
Thus, the values of [tex]\( x \)[/tex] where [tex]\(\frac{x^2 + 2x + 3}{x^2 - x - 12}\)[/tex] is discontinuous are [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
Among the choices given:
- [tex]\( x = -1 \)[/tex]
- [tex]\( x = -2 \)[/tex]
- [tex]\( x = -3 \)[/tex]
- [tex]\( x = -4 \)[/tex]
The value from this list that is a discontinuity for the function is:
[tex]\[ x = -3 \][/tex]
Therefore, the correct answer is:
[tex]\[ x = -3 \][/tex]