{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# Computing k 1 is left as an exercise 1 1 y k x as

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eﬂect across y = x −− − − − − − − − − − − − − −→ switch x and y coordinates (4, −2) y = f −1 ( x ) ? We see that the line x = 4 intersects the graph of the supposed inverse twice - meaning the graph fails the Vertical Line Test, Theorem 1.1, and as such, does not represent y as a function of x. The vertical line x = 4 on the graph on the right corresponds to the horizontal line y = 4 on the graph of y = f (x). The fact that the horizontal line y = 4 intersects the graph of f twice means two diﬀerent inputs, namely x = −2 and x = 2, are matched with the same output, 4, which is the cause of all of the trouble. In general, for a function to have an inverse, diﬀerent inputs must go to diﬀerent outputs, or else we will run into the same problem we did with f (x) = x2 . We give this property a name. Definition 5.3. A function f is said to be one-to-one if f matches diﬀerent inputs to diﬀerent outputs. Equivalently, f is one-to-one if and only if whenever f (c) = f (d), then c = d. Graphically, we detect one-to-one functions using the test below. Theorem 5.4....
View Full Document

{[ snackBarMessage ]}