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have j (x) = j1 (−x). Putting it all together, we have j (x) = j1 (−x) = f (−x + 3) = −x + 3,
5 Or divide - it amounts to the same thing. 92 Relations and Functions
which is what we want.6 From the graph, we conﬁrm the domain of j is (−∞, 3] and we get
the range is [0, ∞). y y (1, 2)
2 2 (−2, 1) (2, 1)
(−1, 2) 1 (3, 0) (−3, 0)
−4 −3 −2 −1 1 y = j1 (x) = √ 2 3 4 x reﬂect across y -axis −− − − − −→
−−−−−− multiply each x-coordinate by −1 x+3 −4 −3 −2 −1 1 2 y = j (x) = j1 (−x) = 3 √ 4 x −x + 3 √
3. The domain of m works out to be the domain of f , [0, ∞). Rewriting m(x) = − x + 3, we
see m(x) = −f (x) + 3. Since we are multiplying the output of f by −1 and then adding
3, we once again have two transformations to deal with: a reﬂection across the x-axis and
a vertical shift. To determine the correct order in which to apply the transformations, we
imagine trying to determine the point on the graph of m which corresponds to (4, 2) on the
graph of f . Since in the formula for m(x), the input to f is just x, we substitute to ﬁnd
m(4) = ...

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