Find solutions to your questions with the help of IDNLearn.com's expert community. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.

(d) Identify any other values of [tex]\(x\)[/tex] (other than those corresponding to vertical asymptotes) for which the function is discontinuous.

[tex]\[ f(x) = \frac{2x + 16}{x^2 + 9x + 8} \][/tex]

is discontinuous at [tex]\( x = \square \)[/tex].


Sagot :

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].