select the correct answer.

which function is increasing and has a domain of $(1, \infty)$?

$f(x) = -\log(x - 2) + 1$

$f(x) = \log(x - 1) + 2$

$f(x) = \log(x - 2) + 1$

$f(x) = -\log(x - 1) + 2$

Question

select the correct answer.

which function is increasing and has a domain of $(1, \infty)$?

$f(x) = -\log(x - 2) + 1$

$f(x) = \log(x - 1) + 2$

$f(x) = \log(x - 2) + 1$

$f(x) = -\log(x - 1) + 2$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Logarithmic Functions", "Function Domain", "Increasing Functions" ], "current_concepts": [ "Logarithmic Functions", "Function Domain", "Increasing Functions" ] } ## Determine the domain of each function $$ \begin{aligned} &\text{For } f(x) = -\log(x - 2) + 1: \quad x - 2 > 0 \implies x \in (2, \infty) \ &\text{For } f(x) = \log(x - 1) + 2: \quad x - 1 > 0 \implies x \in (1, \infty) \ &\text{For } f(x) = \log(x - 2) + 1: \quad x - 2 > 0 \implies x \in (2, \infty) \ &\text{For } f(x) = -\log(x - 1) + 2: \quad x - 1 > 0 \implies x \in (1, \infty) \end{aligned} $$ ## Determine the behavior of the remaining functions $$ \begin{aligned} &f(x) = \log(x - 1) + 2 \implies f'(x) = \frac{1}{(x-1)\ln(10)} > 0 \quad (\text{increasing}) \ &f(x) = -\log(x - 1) + 2 \implies f'(x) = -\frac{1}{(x-1)\ln(10)} < 0 \quad (\text{decreasing}) \end{aligned} $$

f(x) = -\log(x - 2) + 1

f(x) = \log(x - 1) + 2

f(x) = \log(x - 2) + 1

f(x) = -\log(x - 1) + 2

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Logarithmic Functions" ] }

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Question

Explanation:

Determine the domain of each function

Determine the behavior of the remaining functions

Answer: