Answer :
To determine the domain and range of the function [tex]\( f(x) = \log(x-1) + 2 \)[/tex], let's analyze the function step-by-step.
### Domain:
The domain of a function refers to all possible input values [tex]\( x \)[/tex] that will yield a real number output for [tex]\( f(x) \)[/tex].
For the function [tex]\( f(x) = \log(x-1) + 2 \)[/tex]:
1. The logarithm function [tex]\( \log(y) \)[/tex] is defined only for [tex]\( y > 0 \)[/tex]. Therefore, for [tex]\( \log(x-1) \)[/tex] to be defined, the argument [tex]\( x-1 \)[/tex] must be positive.
[tex]\[ x - 1 > 0 \quad \Rightarrow \quad x > 1 \][/tex]
2. Thus, the domain of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] is all [tex]\( x \)[/tex] such that [tex]\( x > 1 \)[/tex].
### Range:
The range of a function refers to all possible output values [tex]\( y \)[/tex] of [tex]\( f(x) \)[/tex].
For the function [tex]\( f(x) = \log(x-1) + 2 \)[/tex]:
1. The range of the logarithmic function [tex]\( \log(y) \)[/tex] itself is all real numbers, i.e., [tex]\( \log(y) \)[/tex] can produce any real number from [tex]\(-\infty\)[/tex] to [tex]\(+\infty\)[/tex].
2. Since adding a constant (in this case, adding 2) to the logarithmic function [tex]\( \log(x-1) \)[/tex] simply shifts its graph vertically, the range of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] remains the set of all real numbers. This vertical shift does not restrict the output values: it only relocates them.
Thus, the range of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] is all real numbers, i.e., [tex]\( (-\infty, \infty) \)[/tex].
### Conclusion:
- The domain of [tex]\( f(x) \)[/tex] is [tex]\( x > 1 \)[/tex].
- The range of [tex]\( f(x) \)[/tex] is all real numbers.
Therefore, the correct choice is:
Domain: [tex]\( x > 1 \)[/tex]; Range: all real numbers
### Domain:
The domain of a function refers to all possible input values [tex]\( x \)[/tex] that will yield a real number output for [tex]\( f(x) \)[/tex].
For the function [tex]\( f(x) = \log(x-1) + 2 \)[/tex]:
1. The logarithm function [tex]\( \log(y) \)[/tex] is defined only for [tex]\( y > 0 \)[/tex]. Therefore, for [tex]\( \log(x-1) \)[/tex] to be defined, the argument [tex]\( x-1 \)[/tex] must be positive.
[tex]\[ x - 1 > 0 \quad \Rightarrow \quad x > 1 \][/tex]
2. Thus, the domain of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] is all [tex]\( x \)[/tex] such that [tex]\( x > 1 \)[/tex].
### Range:
The range of a function refers to all possible output values [tex]\( y \)[/tex] of [tex]\( f(x) \)[/tex].
For the function [tex]\( f(x) = \log(x-1) + 2 \)[/tex]:
1. The range of the logarithmic function [tex]\( \log(y) \)[/tex] itself is all real numbers, i.e., [tex]\( \log(y) \)[/tex] can produce any real number from [tex]\(-\infty\)[/tex] to [tex]\(+\infty\)[/tex].
2. Since adding a constant (in this case, adding 2) to the logarithmic function [tex]\( \log(x-1) \)[/tex] simply shifts its graph vertically, the range of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] remains the set of all real numbers. This vertical shift does not restrict the output values: it only relocates them.
Thus, the range of [tex]\( f(x) = \log(x-1) + 2 \)[/tex] is all real numbers, i.e., [tex]\( (-\infty, \infty) \)[/tex].
### Conclusion:
- The domain of [tex]\( f(x) \)[/tex] is [tex]\( x > 1 \)[/tex].
- The range of [tex]\( f(x) \)[/tex] is all real numbers.
Therefore, the correct choice is:
Domain: [tex]\( x > 1 \)[/tex]; Range: all real numbers