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**Unformatted text preview: **s y = 0. Next, we subtract 3 from the y -coordinates, shifting the graph 6.1 Introduction to Exponential and Logarithmic Functions 339 down 3 units. We get the points 0, − 5 , (1, −2) and (2, −1) with the horizontal asymptote
2
now at y = −3. Connecting the dots in the order and manner as they were on the graph of
g , we get the graph below. We see that the domain of f is the same as g , namely (−∞, ∞),
but that the range of f is (−3, ∞).
y y
8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 −3 −2 −1
−1 1 2 3 4 x −3 −2 −1
−1 1 2 3 4 x −2 −2
−3 y = h(x) = 2x −− − − − −→
−−−−−− y = f (x) = 2x−1 − 3 2. The graph of f passes the Horizontal Line Test so f is one-to-one, hence invertible. To ﬁnd
a formula for f −1 (x), we normally set y = f (x), interchange the x and y , then proceed to
solve for y . Doing so in this situation leads us to the equation x = 2y−1 − 3. We have yet
to discuss how to solve this kind of equation, so we will attempt to ﬁnd the formula for f...

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