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College Algebra, 2013a

College Algebra, 2013a

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364 Further Topics in Functions<br />

=<br />

=<br />

=<br />

=<br />

=<br />

( ) 2x<br />

2<br />

x +1<br />

( ) 2x<br />

+1<br />

x +1<br />

4x<br />

x +1 (x +1)<br />

2x (x +1)<br />

+1·<br />

x +1<br />

4x<br />

x+1<br />

· (x +1)<br />

( 2x<br />

x +1<br />

4x<br />

✘ ✘ ✘✘✿ 1 ✘✘✘✘<br />

(x +1)<br />

(x +1)· 2x<br />

✘ ✘✘ ✘✿ 1✘ (x<br />

(x +1)· ✘ +1)+x ✘✘<br />

+1<br />

4x<br />

3x +1<br />

)<br />

· (x +1)+1· (x +1)<br />

ˆ outside in: This approach yields<br />

(h ◦ h)(x) = h(h(x)) = 2(h(x))<br />

h(x)+1<br />

( ) 2x<br />

2<br />

x +1<br />

= ( ) 2x<br />

+1<br />

x +1<br />

4x<br />

=<br />

3x +1<br />

same algebra as before<br />

To find the domain of h ◦ h, we analyze<br />

(h ◦ h)(x) =<br />

( 2x<br />

2<br />

x +1<br />

( 2x<br />

x +1<br />

)<br />

)<br />

+1<br />

To keep the denominator x + 1 happy, we need x ≠ −1. Setting the denominator<br />

2x<br />

x +1 +1=0<br />

gives x = − 1 3 . Our domain is (−∞, −1) ∪ ( −1, − 1 3)<br />

∪<br />

(<br />

−<br />

1<br />

3 , ∞) .

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