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Which statement could be used to explain why [tex]\( f(x) = 2x - 3 \)[/tex] has an inverse relation that is a function?

A. The graph of [tex]\( f(x) \)[/tex] passes the vertical line test.
B. [tex]\( f(x) \)[/tex] is a one-to-one function.
C. The graph of the inverse of [tex]\( f(x) \)[/tex] passes the horizontal line test.
D. [tex]\( f(x) \)[/tex] is not a function.


Sagot :

To explain why [tex]\( f(x) = 2x - 3 \)[/tex] has an inverse relation that is a function, let's analyze the properties of the function.

First, recall that for a function to have an inverse that is also a function, the original function must be one-to-one. A one-to-one function is defined as a function where each value of [tex]\( x \)[/tex] produces a unique value of [tex]\( f(x) \)[/tex]. In other words, no two different input values should map to the same output value.

The function given is [tex]\( f(x) = 2x - 3 \)[/tex]. To verify if this function is one-to-one, we can check if it passes the Horizontal Line Test. This test states that a function is one-to-one if and only if every horizontal line intersects the graph of the function at most once.

Furthermore, another way to determine if [tex]\( f(x) \)[/tex] is one-to-one is to look at its derivatives. For a linear function [tex]\( f(x) = 2x - 3 \)[/tex]:

1. Compute the derivative: [tex]\( f'(x) = 2 \)[/tex]
2. Since the derivative [tex]\( f'(x) = 2 \)[/tex] is a constant positive value, it indicates that the function is strictly increasing. A strictly increasing function means that for any two different input values [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] where [tex]\( x_1 \neq x_2 \)[/tex], we have [tex]\( f(x_1) \neq f(x_2) \)[/tex].

Given all these explanations, we conclude that [tex]\( f(x) = 2x - 3 \)[/tex] is indeed a one-to-one function. Therefore, this function does have an inverse that is also a function.

Hence, the correct statement to explain why [tex]\( f(x) = 2x - 3 \)[/tex] has an inverse relation that is a function is:
[tex]\[ \boxed{f(x) \text{ is a one-to-one function.}} \][/tex]