Answer :
To determine the domain of the function [tex]\( f(x) = \frac{x+1}{x^2 - 6x + 8} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] for which the function is undefined. The function [tex]\( f(x) \)[/tex] will be undefined where the denominator is zero, as division by zero is not allowed in mathematics.
So, the first step is to find the values of [tex]\( x \)[/tex] for which the denominator [tex]\( x^2 - 6x + 8 \)[/tex] equals zero. To do this, we solve the equation:
[tex]\[ x^2 - 6x + 8 = 0 \][/tex]
This is a quadratic equation and can be factored. We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. Therefore, we can factor the equation as:
[tex]\[ (x - 2)(x - 4) = 0 \][/tex]
Next, we set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
These solutions tell us the values of [tex]\( x \)[/tex] where the denominator is zero, specifically [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex]. Hence, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex].
Thus, the domain of [tex]\( f(x) \)[/tex] includes all real numbers except [tex]\( 2 \)[/tex] and [tex]\( 4 \)[/tex]. Therefore, the domain of the function can be expressed as:
[tex]\[ \text{All real numbers except } 2 \text{ and } 4 \][/tex]
So, the correct answer is:
[tex]\[ \text{All real numbers except 2 and 4} \][/tex]
So, the first step is to find the values of [tex]\( x \)[/tex] for which the denominator [tex]\( x^2 - 6x + 8 \)[/tex] equals zero. To do this, we solve the equation:
[tex]\[ x^2 - 6x + 8 = 0 \][/tex]
This is a quadratic equation and can be factored. We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. Therefore, we can factor the equation as:
[tex]\[ (x - 2)(x - 4) = 0 \][/tex]
Next, we set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
These solutions tell us the values of [tex]\( x \)[/tex] where the denominator is zero, specifically [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex]. Hence, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex].
Thus, the domain of [tex]\( f(x) \)[/tex] includes all real numbers except [tex]\( 2 \)[/tex] and [tex]\( 4 \)[/tex]. Therefore, the domain of the function can be expressed as:
[tex]\[ \text{All real numbers except } 2 \text{ and } 4 \][/tex]
So, the correct answer is:
[tex]\[ \text{All real numbers except 2 and 4} \][/tex]
