Answer :
To identify values of [tex]\( x \)[/tex] at which the function [tex]\( f(x) = \frac{2x + 16}{x^2 + 9x + 8} \)[/tex] is discontinuous, we need to examine where the denominator is zero since a function is discontinuous where its denominator is zero.
Step-by-step solution:
1. Identify the Denominator of the Function:
[tex]\[ f(x) = \frac{2x + 16}{x^2 + 9x + 8} \][/tex]
The denominator is [tex]\( x^2 + 9x + 8 \)[/tex].
2. Find the Roots of the Denominator:
To find where the denominator is zero, solve the equation:
[tex]\[ x^2 + 9x + 8 = 0 \][/tex]
3. Solve the Quadratic Equation:
The equation [tex]\( x^2 + 9x + 8 = 0 \)[/tex] is a quadratic equation. Factoring the quadratic expression, we get:
[tex]\[ (x + 8)(x + 1) = 0 \][/tex]
4. Find the Values of [tex]\( x \)[/tex] That Make the Denominator Zero:
Set each factor equal to zero:
[tex]\[ x + 8 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
Solving these equations gives:
[tex]\[ x = -8 \quad \text{or} \quad x = -1 \][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] is discontinuous at the values of [tex]\( x \)[/tex] where the denominator is zero. These values are:
[tex]\[ x = -8 \quad \text{and} \quad x = -1 \][/tex]
In conclusion:
[tex]\( f(x) = \frac{2x + 16}{x^2 + 9x + 8} \)[/tex] is discontinuous at [tex]\( x = -8 \)[/tex] and [tex]\( x = -1 \)[/tex].
Step-by-step solution:
1. Identify the Denominator of the Function:
[tex]\[ f(x) = \frac{2x + 16}{x^2 + 9x + 8} \][/tex]
The denominator is [tex]\( x^2 + 9x + 8 \)[/tex].
2. Find the Roots of the Denominator:
To find where the denominator is zero, solve the equation:
[tex]\[ x^2 + 9x + 8 = 0 \][/tex]
3. Solve the Quadratic Equation:
The equation [tex]\( x^2 + 9x + 8 = 0 \)[/tex] is a quadratic equation. Factoring the quadratic expression, we get:
[tex]\[ (x + 8)(x + 1) = 0 \][/tex]
4. Find the Values of [tex]\( x \)[/tex] That Make the Denominator Zero:
Set each factor equal to zero:
[tex]\[ x + 8 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
Solving these equations gives:
[tex]\[ x = -8 \quad \text{or} \quad x = -1 \][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] is discontinuous at the values of [tex]\( x \)[/tex] where the denominator is zero. These values are:
[tex]\[ x = -8 \quad \text{and} \quad x = -1 \][/tex]
In conclusion:
[tex]\( f(x) = \frac{2x + 16}{x^2 + 9x + 8} \)[/tex] is discontinuous at [tex]\( x = -8 \)[/tex] and [tex]\( x = -1 \)[/tex].
