Answer :
To find the domain of the function [tex]\( f(x) = \frac{20}{x^2 - 5x + 6} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which the function is defined. The function [tex]\( f(x) \)[/tex] is defined for all [tex]\( x \)[/tex] except where the denominator is zero because division by zero is undefined.
1. Identify the denominator:
The denominator of the function is [tex]\( x^2 - 5x + 6 \)[/tex].
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
3. Factor the quadratic equation:
To factor [tex]\( x^2 - 5x + 6 \)[/tex], we look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.
Hence, we can factor the quadratic as:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
4. Solve the factored equation:
Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
5. Determine the domain:
The values [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex] are the roots where the denominator is zero, and thus, [tex]\( f(x) \)[/tex] is undefined at these values. However, [tex]\( f(x) \)[/tex] is defined for all other real numbers.
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except 2 and 3. In interval notation, this is:
[tex]\[ (-\infty, 2) \cup (2, 3) \cup (3, \infty) \][/tex]
So, the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, 2) \cup (2, 3) \cup (3, \infty) \][/tex]
1. Identify the denominator:
The denominator of the function is [tex]\( x^2 - 5x + 6 \)[/tex].
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
3. Factor the quadratic equation:
To factor [tex]\( x^2 - 5x + 6 \)[/tex], we look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.
Hence, we can factor the quadratic as:
[tex]\[ x^2 - 5x + 6 = (x - 2)(x - 3) \][/tex]
4. Solve the factored equation:
Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
5. Determine the domain:
The values [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex] are the roots where the denominator is zero, and thus, [tex]\( f(x) \)[/tex] is undefined at these values. However, [tex]\( f(x) \)[/tex] is defined for all other real numbers.
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers except 2 and 3. In interval notation, this is:
[tex]\[ (-\infty, 2) \cup (2, 3) \cup (3, \infty) \][/tex]
So, the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, 2) \cup (2, 3) \cup (3, \infty) \][/tex]
