**Unformatted text preview: **d (âˆ’1, 2). If you are experiencing dÂ´j` vu, there is a good
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reason for it but we leave it to the reader to determine the source of this uncanny familiarity.
We obtain the graph below. The domain of f âˆ’1 is (âˆ’3, âˆž), which matches the range of f ,
and the range of f âˆ’1 is (âˆ’âˆž, âˆž), which matches the domain of f .
y y
4 4 3 3 2 2 1 1 âˆ’3 âˆ’2 âˆ’1
âˆ’1 1 2 3 4 5 6 7 8 x âˆ’2 âˆ’1
âˆ’1 âˆ’2
âˆ’3 1 2 3 4 5 6 7 8 x âˆ’2
âˆ’3 y = j (x) = log2 (x) âˆ’âˆ’ âˆ’ âˆ’ âˆ’ âˆ’â†’
âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’ y = f âˆ’1 (x) = log2 (x + 3) + 1 4. We now verify that f (x) = 2xâˆ’1 âˆ’ 3 and f âˆ’1 (x) = log2 (x + 3) + 1 satisfy the composition
requirement for inverses. For all real numbers x, f âˆ’1 â—¦ f (x) = f âˆ’1 (f (x))
= f âˆ’1 2xâˆ’1 âˆ’ 3
= log2 2xâˆ’1 âˆ’ 3 + 3 + 1 = log2 2xâˆ’1 + 1
= (x âˆ’ 1) + 1 Since log2 (2u ) = u for all real numbers u =x
For all real numbers x > âˆ’3, we have9 f â—¦ f âˆ’1 (x) = f f âˆ’1 (x)
= f (log2 (x + 3) + 1)
= 2(log2 (x+3)+1)âˆ’1 âˆ’ 3
= 2log2 (x+3) âˆ...

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