Answer :
To find the domain and range of the function [tex]\( f(x) = -\log(5-x) + 9 \)[/tex], let's go through the steps in detail.
### Domain:
The domain of a function is the set of all possible input values (x-values) that the function can accept. For the logarithmic function [tex]\(\log(5-x)\)[/tex], we need to consider where the log function is defined.
1. The logarithmic function [tex]\(\log(z)\)[/tex] is only defined for [tex]\(z > 0\)[/tex].
2. In our function, this means [tex]\(5-x > 0\)[/tex].
3. Solving the inequality [tex]\(5 - x > 0\)[/tex]:
[tex]\[ 5 - x > 0 \][/tex]
[tex]\[ -x > -5 \][/tex]
[tex]\[ x < 5 \][/tex]
So, the domain of [tex]\(f(x)\)[/tex] is [tex]\(x < 5\)[/tex].
### Range:
The range of a function is the set of all possible output values (y-values) that the function can produce.
1. Consider the inner function [tex]\( \log(5-x) \)[/tex]:
- Since [tex]\(5 - x\)[/tex] is always positive and less than 5, [tex]\(\log(5-x)\)[/tex] can take any real value, from negative to positive infinity.
2. The function [tex]\(f(x)\)[/tex] is given by [tex]\( -\log(5-x) + 9 \)[/tex].
3. Now, consider the negation [tex]\(-\log(5-x)\)[/tex]. When [tex]\(\log(5-x)\)[/tex] is large positive, [tex]\(-\log(5-x)\)[/tex] becomes large negative and when [tex]\(\log(5-x)\)[/tex] is large negative, [tex]\(-\log(5-x)\)[/tex] becomes large positive.
4. Adding 9 shifts the entire range of [tex]\(-\log(5-x)\)[/tex] up by 9 units.
Thus, the minimum value of [tex]\( -\log(5-x) \)[/tex] occurs when [tex]\(\log(5-x)\)[/tex] takes its maximum value (which is when [tex]\( 5-x \)[/tex] approaches 0 from the right, i.e., [tex]\(x \)[/tex] approaches 5 from the left), making [tex]\(-\log(5-x)\)[/tex] very negative.
Therefore, [tex]\( y = -\log(5-x) + 9 \)[/tex] can take on values [tex]\(y\)[/tex] such that [tex]\(y \geq 9\)[/tex].
### Conclusion:
The correct pair of domain and range is:
- Domain: [tex]\(x < 5\)[/tex]
- Range: [tex]\(y \geq 9\)[/tex]
### Domain:
The domain of a function is the set of all possible input values (x-values) that the function can accept. For the logarithmic function [tex]\(\log(5-x)\)[/tex], we need to consider where the log function is defined.
1. The logarithmic function [tex]\(\log(z)\)[/tex] is only defined for [tex]\(z > 0\)[/tex].
2. In our function, this means [tex]\(5-x > 0\)[/tex].
3. Solving the inequality [tex]\(5 - x > 0\)[/tex]:
[tex]\[ 5 - x > 0 \][/tex]
[tex]\[ -x > -5 \][/tex]
[tex]\[ x < 5 \][/tex]
So, the domain of [tex]\(f(x)\)[/tex] is [tex]\(x < 5\)[/tex].
### Range:
The range of a function is the set of all possible output values (y-values) that the function can produce.
1. Consider the inner function [tex]\( \log(5-x) \)[/tex]:
- Since [tex]\(5 - x\)[/tex] is always positive and less than 5, [tex]\(\log(5-x)\)[/tex] can take any real value, from negative to positive infinity.
2. The function [tex]\(f(x)\)[/tex] is given by [tex]\( -\log(5-x) + 9 \)[/tex].
3. Now, consider the negation [tex]\(-\log(5-x)\)[/tex]. When [tex]\(\log(5-x)\)[/tex] is large positive, [tex]\(-\log(5-x)\)[/tex] becomes large negative and when [tex]\(\log(5-x)\)[/tex] is large negative, [tex]\(-\log(5-x)\)[/tex] becomes large positive.
4. Adding 9 shifts the entire range of [tex]\(-\log(5-x)\)[/tex] up by 9 units.
Thus, the minimum value of [tex]\( -\log(5-x) \)[/tex] occurs when [tex]\(\log(5-x)\)[/tex] takes its maximum value (which is when [tex]\( 5-x \)[/tex] approaches 0 from the right, i.e., [tex]\(x \)[/tex] approaches 5 from the left), making [tex]\(-\log(5-x)\)[/tex] very negative.
Therefore, [tex]\( y = -\log(5-x) + 9 \)[/tex] can take on values [tex]\(y\)[/tex] such that [tex]\(y \geq 9\)[/tex].
### Conclusion:
The correct pair of domain and range is:
- Domain: [tex]\(x < 5\)[/tex]
- Range: [tex]\(y \geq 9\)[/tex]
