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**Unformatted text preview: **(h ◦ g )(x) = 2−√x+3 +1 . To
(
) Hence, (h ◦ g )(x) =
simpliﬁcation: domain of h ◦ g , we look to the step before any keep the square root happy, we require x + 3 ≥ 0
√
√
or x ≥ −3. Setting the denominator equal to zero gives 2 − x + 3 + 1 = 0 or x + 3 = 3.
Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation,
√
2 − x + 3 + 1 = 0, we know x = 6 is the only zero of the denominator. Hence, the domain
of h ◦ g is [−3, 6) ∪ (6, ∞).
5. To ﬁnd (h ◦ h)(x), we substitute the function h into itself, h(h(x)).
• inside out : We insert the expression h(x) into h to get
(h ◦ h)(x) = h(h(x)) = h 2x
x+1 2x
x+1
2x
+1
x+1
4x
x + 1 · (x + 1)
2x
(x + 1)
+1
x+1
2 = = 284 Further Topics in Functions
4x
· (x + 1)
x+1 = 2x
· (x + 1) + 1 · (x + 1)
x+1
4x
$
· (x$$
+ 1)
X
$
$ 1$
(x +
$$ 1)
2x
$
+ 1)
· (x$$ + x + 1
X$
$$ 1
(x +
$$ 1)
4x
3x + 1 = = • outside in : This approach yields
(h ◦ h)(x) = h(h(x)) = 2(h(x))
h(x) + 1
2x
x+1
2x
+1
x+1
4x
3x + 1
2 = = same algebra as...

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