Answer :
To find the domain of the function [tex]\( f(x) = \frac{1}{(x-3)(x-5)} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] for which the function is undefined.
The function [tex]\( f(x) \)[/tex] is a rational function, and it is undefined wherever the denominator is zero. Therefore, we need to find the values of [tex]\( x \)[/tex] that make the denominator equal to zero:
1. Consider the denominator [tex]\((x-3)(x-5)\)[/tex].
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x-3)(x-5) = 0 \][/tex]
3. To find the zeros, solve each factor separately:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x - 5 = 0 \][/tex]
4. Solve these equations:
[tex]\[ x = 3 \quad \text{or} \quad x = 5 \][/tex]
Therefore, the function [tex]\( f(x) = \frac{1}{(x-3)(x-5)} \)[/tex] is undefined at [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
Hence, the domain of the function is all real numbers except [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
The correct answer is:
All real numbers except 3 and 5
The function [tex]\( f(x) \)[/tex] is a rational function, and it is undefined wherever the denominator is zero. Therefore, we need to find the values of [tex]\( x \)[/tex] that make the denominator equal to zero:
1. Consider the denominator [tex]\((x-3)(x-5)\)[/tex].
2. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x-3)(x-5) = 0 \][/tex]
3. To find the zeros, solve each factor separately:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x - 5 = 0 \][/tex]
4. Solve these equations:
[tex]\[ x = 3 \quad \text{or} \quad x = 5 \][/tex]
Therefore, the function [tex]\( f(x) = \frac{1}{(x-3)(x-5)} \)[/tex] is undefined at [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
Hence, the domain of the function is all real numbers except [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
The correct answer is:
All real numbers except 3 and 5