Consider the tables that represent ordered pairs corresponding to a function and its inverse. When comparing the functions using the values in the table, which conclusion can be made? According to the tables, f (x) does not have a y-intercept. According to the tables, f-1 (x) does not have an x-intercept. The domain of f (x) is restricted such that x ≥ 0, so the domain of f-1 (x) is restricted such that y ≥ 0. The range of f (x) includes values such that y ≥ 1, so the domain of f-1 (x) includes values such that x ≥ 1.

Question

Lukas Espinoza

Mathematics

5 March, 03:46

Consider the tables that represent ordered pairs corresponding to a function and its inverse. When comparing the functions using the values in the table, which conclusion can be made? According to the tables, f (x) does not have a y-intercept. According to the tables, f-1 (x) does not have an x-intercept. The domain of f (x) is restricted such that x ≥ 0, so the domain of f-1 (x) is restricted such that y ≥ 0. The range of f (x) includes values such that y ≥ 1, so the domain of f-1 (x) includes values such that x ≥ 1.

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Babykins · Answer 1

D) The range of f (x) includes values such that y ≥ 1, so the domain of f-1 (x) includes values such that x ≥ 1.

Step-by-step explanation:

The missing tables are:

First table

x: 0 1 2

f (x) : 1 10 100

Second table

x: 1000 100 10

f^-1 (x) : 3 2 1

Option A is not correct because f (x) has a y-intercept at (0, 1)

If f (x) has a y-intercept, then f^-1 (x) has a x-intercept, which is located at (1, 0). Then option B is not correct

Option C is not correct because the domain of f^-1 (x) is associated with x values.

Option D is correct because the domain of f (x) is the range of f^-1 (x) and vice versa