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**Unformatted text preview: **side out : We insert the expression g (x) into f ﬁrst to get
(f ◦ g )(x) =
=
=
=
=
= f (g (x))
√
f 2− x+3
√
√
2
2− x+3 −4 2− x+3
√
√
√
2
4−4 x+3+
x+3 −8+4 x+3
4+x+3−8
x−1 • outside in : We use the formula for f (x) ﬁrst to get
(f ◦ g )(x) =
=
=
= f (g (x))
(g (x))2 − 4 (g (x))
√
√
2
2− x+3 −4 2− x+3
x−1 same algebra as before Thus we get (f ◦ g )(x) = x − 1. To ﬁnd the domain of (f ◦ g ), we look to the step before
√
√
2
we did any simpliﬁcation and ﬁnd (f ◦ g )(x) = 2 − x + 3 − 4 2 − x + 3 . To keep the
square root happy, we set x + 3 ≥ 0 and ﬁnd our domain to be [−3, ∞).
3. To ﬁnd (g ◦ h)(x), we compute g (h(x)). 282 Further Topics in Functions
• inside out : We insert the expression h(x) into g ﬁrst to get
(g ◦ h)(x) = g (h(x)) = g = 2− = 2− 2x
3(x + 1)
+
x+1
x+1 = 2− 5x + 3
x+1 2x
x+1
2x
x+1 +3
get common denominators • outside in : We use the formula for g (x) ﬁrst to get
(g ◦ h)(x) =
= g (h(x))
2 − h(x) + 3 = 2− 2x
x+1 = 2− 5x +...

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