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**Unformatted text preview: **ossibility, but we also get the possibility that c = 2 − d. This
suggests that f may not be one-to-one. Taking d = 0, we get c = 0 or c = 2. With
f (0) = 4 and f (2) = 4, we have produced two diﬀerent inputs with the same output
meaning f is not one-to-one.
(b) We note that h is a quadratic function and we graph y = h(x) using the techniques
presented in Section 2.3. The vertex is (1, 3) and the parabola opens upwards. We see
immediately from the graph that h is not one-to-one, since there are several horizontal
lines which cross the graph more than once.
4. (a) The function F is given to us as a set of ordered pairs. The condition F (c) = F (d)
means the outputs from the function (the y -coordinates of the ordered pairs) are the
same. We see that the points (−1, 1) and (2, 1) are both elements of F with F (−1) = 1
and F (2) = 1. Since −1 = 2, we have established that F is not one-to-one.
(b) Graphically, we see the horizontal line y = 1 crosses the graph more than once. Hence,
the graph of F fails the Horizontal Line Test. 5.2 Inverse Functions 299 y
6 y 5
4 2 3 1 x
2
−2 −1 1 2 1 y = F (x)
1 2 x −1 y = h(x) We have shown that the functions f and g in Example 5.2.1 are one-to-one. This means they are
invertible, so it i...

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