Question

In Exploration 1.2.2 you worked on understanding increasing and decreasing functions. Question 2 asked if the graph is increasing on [2,4].

Which of the following is deemed the strongest answer?

Option 1: It is both increasing and decreasing because when you graph the function, part of the function goes up and part of the function goes down.
Option 2: The function is decreasing because 2<2.5 and f(2)>f(2.5) , therefore by the defnition it is decreasing.
Option 3: The function is increasing because 2<3 and f(2) AND 2<4 and f(2) , therefore by the defnition it is increasing.
Option 4: The function is increasing because 2<4 and f(2) , therefore by the defnition it is increasing.
Option 5: It is both increasing and decreasing in the interval [2,4] , because 2<2.5 and f(2)>f(2.5) AND 3<4 and f(3) .

Answer 1

The function is increasing because 2<4 and f(2), therefore by the definition it is increasing.Option 4 is the correct answer.

Step-by-step explanation:

The answer is Option 4: "The function is increasing because 2 < 4 and f(2)." This choice aligns with the fundamental definition of an increasing function. According to this definition, if for any two points (a) and (b) in the domain where (a < b), f(a) ≤ f(b), the function is considered increasing over that interval. Option 4 correctly identifies that as (2 < 4), and (f(2) < f(4)), the function is increasing on the interval [2, 4].

Option 2, though providing a statement about the function being decreasing, contains an error in reasoning, stating "2 < 2.5 and f(2) > f(2.5)." This violates the definition of a decreasing function, as it suggests that the function is increasing between 2 and 2.5, which contradicts the claim of it being decreasing.

Option 5 is also weakened by its contradiction, claiming both increasing and decreasing within the same interval. Such an assertion conflicts with the basic principles of function behavior on specific intervals. Thus, Option 4 stands out as the most accurate and logically sound choice based on the provided information.