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

**Unformatted text preview: **piciously
like some kind of distributive property, it is nothing of the sort; the addition on the left hand side
of the equation is function addition, and we are using this equation to deﬁne the output of the
new function f + g as the sum of the real number outputs from f and g .1
Example 1.6.1. Let f (x) = 6x2 − 2x and g (x) = 3 − 1
. Find and simplify expressions for
x 1. (f + g )(x) 3. (f g )(x) 2. (g − f )(x) 4. 1 g
f (x) The author is well aware that this point is pedantic, and lost on most readers. 56 Relations and Functions In addition, ﬁnd the domain of each of these functions.
Solution.
1. (f + g )(x) is deﬁned to be f (x) + g (x). To that end, we get (f + g )(x) = f (x) + g (x)
= 6x2 − 2x + 3 − 1
x 1
x
6x3 2x2 3x 1
−
+
−
x
x
x
x
3 − 2x2 + 3x − 1
6x
x = 6 x2 − 2x + 3 −
=
= get common denominators To ﬁnd the domain of (f + g ) we do so before we simplify, that is, at the step
6x2 − 2x + 3 − 1
x We see x = 0, but everything else is ﬁne. Hence, the domain is (−∞, 0) ∪ (0, ∞).
2. (g...

View
Full
Document