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

**Unformatted text preview: **(Why?) Graphically, The graph of h appears to be neither symmetric about the y -axis nor the origin.
4.
i(x) =
i(−x) =
i(−x) = 5x
2x − x3
5(−x)
2(−x) − (−x)3
−5x
−2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some
x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in
fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however, 1.7 Graphs of Functions 69 prove i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x)
and i(−x) and show they are equivalent. A clue as to how to proceed is in the numerators:
in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To re-write
i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value
of the fraction the same, we need to multiply the denominator by −1 as well. Thus
i(x) =
=
= 5x
2x − x3
(−1)5x
(−1) (2x − x3 )
−5x
−2x + x3 Hence, i(x) = i...

View
Full
Document